1-DAY-CHEMISTRY | Princy Online

UNIT 1Mole Concept

Some Basic Concepts of ChemistryMole Concept, Determination of Formula of Compound, Stoichiometric Calculations, Concentration Terms, Relation Between Molarity and Normality

1. Determine the number of moles in 20 grams of water (H₂O).

Molecular weight of H2O=2×1+16=18 g/mol\text{Molecular weight of H}_2\text{O} = 2 \times 1 + 16 = 18 \, \text{g/mol}Molecular weight of H2O=2×1+16=18g/mol

Moles of H2O=massmolecular weight=20 g18 g/mol=1.11 mol\text{Moles of H}_2\text{O} = \frac{\text{mass}}{\text{molecular weight}} = \frac{20 \, \text{g}}{18 \, \text{g/mol}} = 1.11 \, \text{mol}Moles of H2O=molecular weightmass=18g/mol20g=1.11mol

Explanation: The molecular weight is calculated using atomic masses. Dividing the mass of the substance by its molecular weight gives the number of moles.

Determination of Formula of Compound

2. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 40 g
Hydrogen: 6.7 g
Oxygen: 53.3 g

Step 2: Convert mass to moles:

Moles of C=40 g12 g/mol=3.33 mol\text{Moles of C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} = 3.33 \, \text{mol}Moles of C=12g/mol40g=3.33mol Moles of H=6.7 g1 g/mol=6.7 mol\text{Moles of H} = \frac{6.7 \, \text{g}}{1 \, \text{g/mol}} = 6.7 \, \text{mol}Moles of H=1g/mol6.7g=6.7mol Moles of O=53.3 g16 g/mol=3.33 mol\text{Moles of O} = \frac{53.3 \, \text{g}}{16 \, \text{g/mol}} = 3.33 \, \text{mol}Moles of O=16g/mol53.3g=3.33mol

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=3.33:6.7:3.33=1:2:1\text{Ratio of C:H:O} = 3.33:6.7:3.33 = 1:2:1Ratio of C:H:O=3.33:6.7:3.33=1:2:1

Empirical formula: CH₂O

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations

3. Calculate the mass of CO₂ produced from the complete combustion of 10 g of methane (CH₄).

CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}CH4+2O2→CO2+2H2O

Step 1: Determine moles of CH₄:

Molecular weight of CH4=12+4×1=16 g/mol\text{Molecular weight of CH}_4 = 12 + 4 \times 1 = 16 \, \text{g/mol}Molecular weight of CH4=12+4×1=16g/mol Moles of CH4=10 g16 g/mol=0.625 mol\text{Moles of CH}_4 = \frac{10 \, \text{g}}{16 \, \text{g/mol}} = 0.625 \, \text{mol}Moles of CH4=16g/mol10g=0.625mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of CO₂:

Moles of CO2=0.625 mol\text{Moles of CO}_2 = 0.625 \, \text{mol}Moles of CO2=0.625mol

Step 3: Calculate the mass of CO₂:

Molecular weight of CO2=12+2×16=44 g/mol\text{Molecular weight of CO}_2 = 12 + 2 \times 16 = 44 \, \text{g/mol}Molecular weight of CO2=12+2×16=44g/mol Mass of CO2=0.625 mol×44 g/mol=27.5 g\text{Mass of CO}_2 = 0.625 \, \text{mol} \times 44 \, \text{g/mol} = 27.5 \, \text{g}Mass of CO2=0.625mol×44g/mol=27.5g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the product.

Concentration Terms

4. Calculate the molarity of a solution prepared by dissolving 5 g of NaCl in 500 mL of solution.

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Moles of NaCl=5 g58.5 g/mol=0.0855 mol\text{Moles of NaCl} = \frac{5 \, \text{g}}{58.5 \, \text{g/mol}} = 0.0855 \, \text{mol}Moles of NaCl=58.5g/mol5g=0.0855mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=moles of solutevolume of solution in liters=0.0855 mol0.5 L=0.171 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.0855 \, \text{mol}}{0.5 \, \text{L}} = 0.171 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.5L0.0855mol=0.171M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

Relation Between Molarity and Normality

5. Find the normality of a 1 M H₂SO₄ solution.

H2SO4→2H++SO42−\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}H2SO4→2H++SO42−

Step 1: Determine the number of equivalents:

Number of H+ions produced=2\text{Number of H}^+ \text{ions produced} = 2Number of H+ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=1 M×2=2 N\text{Normality} = 1 \, \text{M} \times 2 = 2 \, \text{N}Normality=1M×2=2N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

These examples cover key concepts in chemistry, including mole concept, determination of empirical formulas, stoichiometric calculations, concentration terms, and the relationship between molarity and normality. Each example includes a detailed explanation to aid understanding. For a collection of 200 numerical solved questions, each section should be expanded similarly.

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Mole Concept (continued)

6. Calculate the number of molecules in 2 moles of ammonia (NH₃).

Number of molecules=moles×Avogadro’s number\text{Number of molecules} = \text{moles} \times \text{Avogadro's number}Number of molecules=moles×Avogadro’s number Number of molecules=2 mol×6.022×1023 molecules/mol\text{Number of molecules} = 2 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol}Number of molecules=2mol×6.022×1023molecules/mol Number of molecules=1.2044×1024 molecules\text{Number of molecules} = 1.2044 \times 10^{24} \, \text{molecules}Number of molecules=1.2044×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the total number of molecules.

7. Determine the mass of one molecule of water (H₂O).

Mass of one molecule=Molar massAvogadro’s number\text{Mass of one molecule} = \frac{\text{Molar mass}}{\text{Avogadro's number}}Mass of one molecule=Avogadro’s numberMolar mass Molar mass of H2O=18 g/mol\text{Molar mass of H}_2\text{O} = 18 \, \text{g/mol}Molar mass of H2O=18g/mol Mass of one molecule=18 g/mol6.022×1023 molecules/mol\text{Mass of one molecule} = \frac{18 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}}Mass of one molecule=6.022×1023molecules/mol18g/mol Mass of one molecule=2.99×10−23 g\text{Mass of one molecule} = 2.99 \times 10^{-23} \, \text{g}Mass of one molecule=2.99×10−23g

Explanation: Divide the molar mass by Avogadro's number to find the mass of a single molecule.

Determination of Formula of Compound (continued)

8. A compound contains 70% iron and 30% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Iron: 70 g
Oxygen: 30 g

Step 2: Convert mass to moles:

Moles of Fe=70 g55.85 g/mol=1.25 mol\text{Moles of Fe} = \frac{70 \, \text{g}}{55.85 \, \text{g/mol}} = 1.25 \, \text{mol}Moles of Fe=55.85g/mol70g=1.25mol Moles of O=30 g16 g/mol=1.875 mol\text{Moles of O} = \frac{30 \, \text{g}}{16 \, \text{g/mol}} = 1.875 \, \text{mol}Moles of O=16g/mol30g=1.875mol

Step 3: Determine the simplest mole ratio:

Ratio of Fe:O=1.25:1.875=2:3\text{Ratio of Fe:O} = 1.25:1.875 = 2:3Ratio of Fe:O=1.25:1.875=2:3

Empirical formula: Fe₂O₃

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations (continued)

9. How many grams of H₂ are needed to react completely with 10 g of N₂ to form ammonia (NH₃)?

N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2+3H2→2NH3

Step 1: Determine moles of N₂:

Molecular weight of N2=2×14=28 g/mol\text{Molecular weight of N}_2 = 2 \times 14 = 28 \, \text{g/mol}Molecular weight of N2=2×14=28g/mol Moles of N2=10 g28 g/mol=0.357 mol\text{Moles of N}_2 = \frac{10 \, \text{g}}{28 \, \text{g/mol}} = 0.357 \, \text{mol}Moles of N2=28g/mol10g=0.357mol

Step 2: Use the stoichiometric ratio (1:3) to find moles of H₂:

Moles of H2=0.357 mol×3=1.071 mol\text{Moles of H}_2 = 0.357 \, \text{mol} \times 3 = 1.071 \, \text{mol}Moles of H2=0.357mol×3=1.071mol

Step 3: Calculate the mass of H₂:

Molecular weight of H2=2×1=2 g/mol\text{Molecular weight of H}_2 = 2 \times 1 = 2 \, \text{g/mol}Molecular weight of H2=2×1=2g/mol Mass of H2=1.071 mol×2 g/mol=2.142 g\text{Mass of H}_2 = 1.071 \, \text{mol} \times 2 \, \text{g/mol} = 2.142 \, \text{g}Mass of H2=1.071mol×2g/mol=2.142g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the reactant.

Concentration Terms (continued)

10. Calculate the molality of a solution prepared by dissolving 10 g of KCl in 200 g of water.

Molecular weight of KCl=39+35.5=74.5 g/mol\text{Molecular weight of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol}Molecular weight of KCl=39+35.5=74.5g/mol Moles of KCl=10 g74.5 g/mol=0.134 mol\text{Moles of KCl} = \frac{10 \, \text{g}}{74.5 \, \text{g/mol}} = 0.134 \, \text{mol}Moles of KCl=74.5g/mol10g=0.134mol

Mass of solvent (water)=200 g=0.2 kg\text{Mass of solvent (water)} = 200 \, \text{g} = 0.2 \, \text{kg}Mass of solvent (water)=200g=0.2kg Molality=moles of solutemass of solvent in kg=0.134 mol0.2 kg=0.67 m\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.134 \, \text{mol}}{0.2 \, \text{kg}} = 0.67 \, \text{m}Molality=mass of solvent in kgmoles of solute=0.2kg0.134mol=0.67m

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

Relation Between Molarity and Normality (continued)

11. Find the normality of a 0.5 M Na₂CO₃ solution.

Na2CO3→2Na++CO32−\text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-}Na2CO3→2Na++CO32−

Step 1: Determine the number of equivalents:

Number of H+ions CO32−can accept=2\text{Number of H}^+ \text{ions CO}_3^{2-} \text{can accept} = 2Number of H+ions CO32−can accept=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×2=1 N\text{Normality} = 0.5 \, \text{M} \times 2 = 1 \, \text{N}Normality=0.5M×2=1N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

These examples expand on the key concepts, providing more detailed numerical problems and their step-by-step solutions. For a collection of 200 numerical solved questions, each section can be further developed with additional problems and explanations.

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Mole Concept (continued)

12. Calculate the number of atoms in 3 moles of aluminum (Al).

Number of atoms=moles×Avogadro’s number\text{Number of atoms} = \text{moles} \times \text{Avogadro's number}Number of atoms=moles×Avogadro’s number Number of atoms=3 mol×6.022×1023 atoms/mol\text{Number of atoms} = 3 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol}Number of atoms=3mol×6.022×1023atoms/mol Number of atoms=1.8066×1024 atoms\text{Number of atoms} = 1.8066 \times 10^{24} \, \text{atoms}Number of atoms=1.8066×1024atoms

Explanation: Multiply the number of moles by Avogadro's number to find the total number of atoms.

13. Determine the mass of 0.5 moles of sulfuric acid (H₂SO₄).

Molecular weight of H2SO4=2×1+32+4×16=98 g/mol\text{Molecular weight of H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H2SO4=2×1+32+4×16=98g/mol \text{Mass of H}_2\text{SO}_4} = 0.5 \, \text{mol} \times 98 \, \text{g/mol} = 49 \, \text{g}

Explanation: Multiply the number of moles by the molecular weight to find the mass of the substance.

Determination of Formula of Compound (continued)

14. A compound contains 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 54.5 g
Hydrogen: 9.1 g
Oxygen: 36.4 g

Step 2: Convert mass to moles:

Moles of C=54.5 g12 g/mol=4.54 mol\text{Moles of C} = \frac{54.5 \, \text{g}}{12 \, \text{g/mol}} = 4.54 \, \text{mol}Moles of C=12g/mol54.5g=4.54mol Moles of H=9.1 g1 g/mol=9.1 mol\text{Moles of H} = \frac{9.1 \, \text{g}}{1 \, \text{g/mol}} = 9.1 \, \text{mol}Moles of H=1g/mol9.1g=9.1mol Moles of O=36.4 g16 g/mol=2.275 mol\text{Moles of O} = \frac{36.4 \, \text{g}}{16 \, \text{g/mol}} = 2.275 \, \text{mol}Moles of O=16g/mol36.4g=2.275mol

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=4.54:9.1:2.275=2:4:1\text{Ratio of C:H:O} = 4.54:9.1:2.275 = 2:4:1Ratio of C:H:O=4.54:9.1:2.275=2:4:1

Empirical formula: C₂H₄O

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations (continued)

15. How many grams of O₂ are required to react completely with 5 g of C₆H₆ (benzene) to form CO₂ and H₂O?

2C6H6+15O2→12CO2+6H2O2\text{C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2\text{O}2C6H6+15O2→12CO2+6H2O

Step 1: Determine moles of C₆H₆:

Molecular weight of C6H6=6×12+6×1=78 g/mol\text{Molecular weight of C}_6\text{H}_6 = 6 \times 12 + 6 \times 1 = 78 \, \text{g/mol}Molecular weight of C6H6=6×12+6×1=78g/mol Moles of C6H6=5 g78 g/mol=0.0641 mol\text{Moles of C}_6\text{H}_6 = \frac{5 \, \text{g}}{78 \, \text{g/mol}} = 0.0641 \, \text{mol}Moles of C6H6=78g/mol5g=0.0641mol

Step 2: Use the stoichiometric ratio (2:15) to find moles of O₂:

Moles of O2=0.0641 mol C6H6×15 mol O22 mol C6H6=0.481 mol\text{Moles of O}_2 = 0.0641 \, \text{mol C}_6\text{H}_6 \times \frac{15 \, \text{mol O}_2}{2 \, \text{mol C}_6\text{H}_6} = 0.481 \, \text{mol}Moles of O2=0.0641mol C6H6×2mol C6H615mol O2=0.481mol

Step 3: Calculate the mass of O₂:

Molecular weight of O2=2×16=32 g/mol\text{Molecular weight of O}_2 = 2 \times 16 = 32 \, \text{g/mol}Molecular weight of O2=2×16=32g/mol Mass of O2=0.481 mol×32 g/mol=15.4 g\text{Mass of O}_2 = 0.481 \, \text{mol} \times 32 \, \text{g/mol} = 15.4 \, \text{g}Mass of O2=0.481mol×32g/mol=15.4g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the reactant.

Concentration Terms (continued)

16. Calculate the mole fraction of ethanol (C₂H₅OH) in a solution containing 46 g of ethanol and 54 g of water.

Molecular weight of C2H5OH=2×12+6×1+16=46 g/mol\text{Molecular weight of C}_2\text{H}_5\text{OH} = 2 \times 12 + 6 \times 1 + 16 = 46 \, \text{g/mol}Molecular weight of C2H5OH=2×12+6×1+16=46g/mol Moles of C2H5OH=46 g46 g/mol=1 mol\text{Moles of C}_2\text{H}_5\text{OH} = \frac{46 \, \text{g}}{46 \, \text{g/mol}} = 1 \, \text{mol}Moles of C2H5OH=46g/mol46g=1mol

Molecular weight of H2O=18 g/mol\text{Molecular weight of H}_2\text{O} = 18 \, \text{g/mol}Molecular weight of H2O=18g/mol Moles of H2O=54 g18 g/mol=3 mol\text{Moles of H}_2\text{O} = \frac{54 \, \text{g}}{18 \, \text{g/mol}} = 3 \, \text{mol}Moles of H2O=18g/mol54g=3mol

Total moles=1 mol+3 mol=4 mol\text{Total moles} = 1 \, \text{mol} + 3 \, \text{mol} = 4 \, \text{mol}Total moles=1mol+3mol=4mol Mole fraction of C2H5OH=1 mol4 mol=0.25\text{Mole fraction of C}_2\text{H}_5\text{OH} = \frac{1 \, \text{mol}}{4 \, \text{mol}} = 0.25Mole fraction of C2H5OH=4mol1mol=0.25

Explanation: Calculate the mole fraction by dividing the moles of the solute by the total moles in the solution.

Relation Between Molarity and Normality (continued)

17. Determine the normality of a 1.5 M solution of phosphoric acid (H₃PO₄).

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3\text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ions produced=3\text{Number of H}^+ \text{ions produced} = 3Number of H+ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=1.5 M×3=4.5 N\text{Normality} = 1.5 \, \text{M} \times 3 = 4.5 \, \text{N}Normality=1.5M×3=4.5N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

These additional examples further cover essential chemistry concepts, providing detailed numerical problems and their step-by-step solutions. Each section can be expanded with more problems to achieve the target of 200 numerical solved questions with explanations.

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Mole Concept (continued)

18. Calculate the number of moles in 25 grams of potassium chloride (KCl).

Molecular weight of KCl=39+35.5=74.5 g/mol\text{Molecular weight of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol}Molecular weight of KCl=39+35.5=74.5g/mol

Moles of KCl=25 g74.5 g/mol=0.335 mol\text{Moles of KCl} = \frac{25 \, \text{g}}{74.5 \, \text{g/mol}} = 0.335 \, \text{mol}Moles of KCl=74.5g/mol25g=0.335mol

Explanation: Divide the mass of the substance by its molecular weight to find the number of moles.

19. Determine the number of atoms in 0.25 moles of sulfur (S).

Number of atoms=moles×Avogadro’s number\text{Number of atoms} = \text{moles} \times \text{Avogadro's number}Number of atoms=moles×Avogadro’s number Number of atoms=0.25 mol×6.022×1023 atoms/mol\text{Number of atoms} = 0.25 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol}Number of atoms=0.25mol×6.022×1023atoms/mol Number of atoms=1.5055×1023 atoms\text{Number of atoms} = 1.5055 \times 10^{23} \, \text{atoms}Number of atoms=1.5055×1023atoms

Explanation: Multiply the number of moles by Avogadro's number to find the total number of atoms.

Determination of Formula of Compound (continued)

20. A compound contains 28.7% nitrogen and 71.3% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Nitrogen: 28.7 g
Oxygen: 71.3 g

Step 2: Convert mass to moles:

Moles of N=28.7 g14 g/mol=2.05 mol\text{Moles of N} = \frac{28.7 \, \text{g}}{14 \, \text{g/mol}} = 2.05 \, \text{mol}Moles of N=14g/mol28.7g=2.05mol Moles of O=71.3 g16 g/mol=4.46 mol\text{Moles of O} = \frac{71.3 \, \text{g}}{16 \, \text{g/mol}} = 4.46 \, \text{mol}Moles of O=16g/mol71.3g=4.46mol

Step 3: Determine the simplest mole ratio:

Ratio of N:O=2.05:4.46=1:2.18\text{Ratio of N:O} = 2.05:4.46 = 1:2.18Ratio of N:O=2.05:4.46=1:2.18

Step 4: Simplify to nearest whole number ratio:

Ratio of N:O≈1:2\text{Ratio of N:O} \approx 1:2Ratio of N:O≈1:2

Empirical formula: NO₂

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations (continued)

21. How many grams of NaCl are produced when 2 g of Na reacts with excess Cl₂?

2Na+Cl2→2NaCl2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}2Na+Cl2→2NaCl

Step 1: Determine moles of Na:

Molecular weight of Na=23 g/mol\text{Molecular weight of Na} = 23 \, \text{g/mol}Molecular weight of Na=23g/mol Moles of Na=2 g23 g/mol=0.087 mol\text{Moles of Na} = \frac{2 \, \text{g}}{23 \, \text{g/mol}} = 0.087 \, \text{mol}Moles of Na=23g/mol2g=0.087mol

Step 2: Use the stoichiometric ratio (2:2) to find moles of NaCl:

Moles of NaCl=0.087 mol\text{Moles of NaCl} = 0.087 \, \text{mol}Moles of NaCl=0.087mol

Step 3: Calculate the mass of NaCl:

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Mass of NaCl=0.087 mol×58.5 g/mol=5.09 g\text{Mass of NaCl} = 0.087 \, \text{mol} \times 58.5 \, \text{g/mol} = 5.09 \, \text{g}Mass of NaCl=0.087mol×58.5g/mol=5.09g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the product.

Concentration Terms (continued)

22. Calculate the molarity of a solution prepared by dissolving 10 g of acetic acid (CH₃COOH) in 250 mL of solution.

Molecular weight of CH3COOH=2×12+4×1+2×16=60 g/mol\text{Molecular weight of CH}_3\text{COOH} = 2 \times 12 + 4 \times 1 + 2 \times 16 = 60 \, \text{g/mol}Molecular weight of CH3COOH=2×12+4×1+2×16=60g/mol Moles of CH3COOH=10 g60 g/mol=0.167 mol\text{Moles of CH}_3\text{COOH} = \frac{10 \, \text{g}}{60 \, \text{g/mol}} = 0.167 \, \text{mol}Moles of CH3COOH=60g/mol10g=0.167mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=moles of solutevolume of solution in liters=0.167 mol0.25 L=0.668 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.167 \, \text{mol}}{0.25 \, \text{L}} = 0.668 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.25L0.167mol=0.668M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

Relation Between Molarity and Normality (continued)

23. Find the normality of a 0.1 M Ba(OH)₂ solution.

Ba(OH)2→Ba2++2OH−\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-Ba(OH)2→Ba2++2OH−

Step 1: Determine the number of equivalents:

Number of OH−ions produced=2\text{Number of OH}^- \text{ions produced} = 2Number of OH−ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.1 M×2=0.2 N\text{Normality} = 0.1 \, \text{M} \times 2 = 0.2 \, \text{N}Normality=0.1M×2=0.2N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

Additional Numerical Problems with Explanations

24. Calculate the mass of glucose (C₆H₁₂O₆) needed to prepare 1 L of a 0.5 M solution.

Molecular weight of C6H12O6=6×12+12×1+6×16=180 g/mol\text{Molecular weight of C}_6\text{H}_{12}\text{O}_6 = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, \text{g/mol}Molecular weight of C6H12O6=6×12+12×1+6×16=180g/mol Moles of C6H12O6=0.5 mol/L×1 L=0.5 mol\text{Moles of C}_6\text{H}_{12}\text{O}_6 = 0.5 \, \text{mol/L} \times 1 \, \text{L} = 0.5 \, \text{mol}Moles of C6H12O6=0.5mol/L×1L=0.5mol Mass of C6H12O6=0.5 mol×180 g/mol=90 g\text{Mass of C}_6\text{H}_{12}\text{O}_6 = 0.5 \, \text{mol} \times 180 \, \text{g/mol} = 90 \, \text{g}Mass of C6H12O6=0.5mol×180g/mol=90g

Explanation: Multiply the molarity by the volume of the solution to find the number of moles, then multiply by the molecular weight to find the mass.

25. Determine the normality of a 0.25 M HCl solution.

HCl→H++Cl−\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-HCl→H++Cl−

Step 1: Determine the number of equivalents:

Number of H+ions produced=1\text{Number of H}^+ \text{ions produced} = 1Number of H+ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.25 M×1=0.25 N\text{Normality} = 0.25 \, \text{M} \times 1 = 0.25 \, \text{N}Normality=0.25M×1=0.25N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

26. Calculate the number of moles in 500 mg of aspirin (C₉H₈O₄).

Molecular weight of C9H8O4=9×12+8×1+4×16=180 g/mol\text{Molecular weight of C}_9\text{H}_8\text{O}_4 = 9 \times 12 + 8 \times 1 + 4 \times 16 = 180 \, \text{g/mol}Molecular weight of C9H8O4=9×12+8×1+4×16=180g/mol Mass of aspirin=500 mg=0.5 g\text{Mass of aspirin} = 500 \, \text{mg} = 0.5 \, \text{g}Mass of aspirin=500mg=0.5g Moles of C9H8O4=0.5 g180 g/mol=0.00278 mol\text{Moles of C}_9\text{H}_8\text{O}_4 = \frac{0.5 \, \text{g}}{180 \, \text{g/mol}} = 0.00278 \, \text{mol}Moles of C9H8O4=180g/mol0.5g=0.00278mol

Explanation: Convert the mass from milligrams to grams, then divide by the molecular weight to find the number of moles.

These additional examples cover essential chemistry concepts, providing detailed numerical problems and their step-by-step solutions. Each section can be further expanded with more problems to reach the target of 200 numerical solved questions with explanations.

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Mole Concept (continued)

27. Calculate the number of molecules in 10 grams of CO₂.

Molecular weight of CO2=12+2×16=44 g/mol\text{Molecular weight of CO}_2 = 12 + 2 \times 16 = 44 \, \text{g/mol}Molecular weight of CO2=12+2×16=44g/mol

Moles of CO2=10 g44 g/mol=0.227 mol\text{Moles of CO}_2 = \frac{10 \, \text{g}}{44 \, \text{g/mol}} = 0.227 \, \text{mol}Moles of CO2=44g/mol10g=0.227mol

Number of molecules=moles×Avogadro’s number\text{Number of molecules} = \text{moles} \times \text{Avogadro's number}Number of molecules=moles×Avogadro’s number Number of molecules=0.227 mol×6.022×1023 molecules/mol=1.367×1023 molecules\text{Number of molecules} = 0.227 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.367 \times 10^{23} \, \text{molecules}Number of molecules=0.227mol×6.022×1023molecules/mol=1.367×1023molecules

Explanation: Convert mass to moles, then multiply by Avogadro's number to find the number of molecules.

28. How many grams of H₂O are produced from 2 moles of H₂?

2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2+O2→2H2O

Step 1: Use the stoichiometric ratio (2:2) to find moles of H₂O:

Moles of H2O=2 mol\text{Moles of H}_2\text{O} = 2 \, \text{mol}Moles of H2O=2mol

Step 2: Calculate the mass of H₂O:

Molecular weight of H2O=2×1+16=18 g/mol\text{Molecular weight of H}_2\text{O} = 2 \times 1 + 16 = 18 \, \text{g/mol}Molecular weight of H2O=2×1+16=18g/mol Mass of H2O=2 mol×18 g/mol=36 g\text{Mass of H}_2\text{O} = 2 \, \text{mol} \times 18 \, \text{g/mol} = 36 \, \text{g}Mass of H2O=2mol×18g/mol=36g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the product.

Determination of Formula of Compound (continued)

29. A compound contains 40% sulfur and 60% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Sulfur: 40 g
Oxygen: 60 g

Step 2: Convert mass to moles:

Moles of S=40 g32 g/mol=1.25 mol\text{Moles of S} = \frac{40 \, \text{g}}{32 \, \text{g/mol}} = 1.25 \, \text{mol}Moles of S=32g/mol40g=1.25mol Moles of O=60 g16 g/mol=3.75 mol\text{Moles of O} = \frac{60 \, \text{g}}{16 \, \text{g/mol}} = 3.75 \, \text{mol}Moles of O=16g/mol60g=3.75mol

Step 3: Determine the simplest mole ratio:

Ratio of S:O=1.25:3.75=1:3\text{Ratio of S:O} = 1.25:3.75 = 1:3Ratio of S:O=1.25:3.75=1:3

Empirical formula: SO₃

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations (continued)

30. How many grams of CO₂ are produced from 10 g of C₆H₆ (benzene) when it is burned in excess oxygen?

2C6H6+15O2→12CO2+6H2O2\text{C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2\text{O}2C6H6+15O2→12CO2+6H2O

Step 1: Determine moles of C₆H₆:

Molecular weight of C6H6=6×12+6×1=78 g/mol\text{Molecular weight of C}_6\text{H}_6 = 6 \times 12 + 6 \times 1 = 78 \, \text{g/mol}Molecular weight of C6H6=6×12+6×1=78g/mol Moles of C6H6=10 g78 g/mol=0.128 mol\text{Moles of C}_6\text{H}_6 = \frac{10 \, \text{g}}{78 \, \text{g/mol}} = 0.128 \, \text{mol}Moles of C6H6=78g/mol10g=0.128mol

Step 2: Use the stoichiometric ratio (2:12) to find moles of CO₂:

Moles of CO2=0.128 mol C6H6×12 mol CO22 mol C6H6=0.768 mol\text{Moles of CO}_2 = 0.128 \, \text{mol C}_6\text{H}_6 \times \frac{12 \, \text{mol CO}_2}{2 \, \text{mol C}_6\text{H}_6} = 0.768 \, \text{mol}Moles of CO2=0.128mol C6H6×2mol C6H612mol CO2=0.768mol

Step 3: Calculate the mass of CO₂:

Molecular weight of CO2=12+2×16=44 g/mol\text{Molecular weight of CO}_2 = 12 + 2 \times 16 = 44 \, \text{g/mol}Molecular weight of CO2=12+2×16=44g/mol Mass of CO2=0.768 mol×44 g/mol=33.8 g\text{Mass of CO}_2 = 0.768 \, \text{mol} \times 44 \, \text{g/mol} = 33.8 \, \text{g}Mass of CO2=0.768mol×44g/mol=33.8g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the product.

Concentration Terms (continued)

31. Calculate the molality of a solution prepared by dissolving 10 g of NaOH in 500 g of water.

Molecular weight of NaOH=23+16+1=40 g/mol\text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}Molecular weight of NaOH=23+16+1=40g/mol Moles of NaOH=10 g40 g/mol=0.25 mol\text{Moles of NaOH} = \frac{10 \, \text{g}}{40 \, \text{g/mol}} = 0.25 \, \text{mol}Moles of NaOH=40g/mol10g=0.25mol

Mass of solvent (water)=500 g=0.5 kg\text{Mass of solvent (water)} = 500 \, \text{g} = 0.5 \, \text{kg}Mass of solvent (water)=500g=0.5kg Molality=moles of solutemass of solvent in kg=0.25 mol0.5 kg=0.5 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.25 \, \text{mol}}{0.5 \, \text{kg}} = 0.5 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.5kg0.25mol=0.5mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

Relation Between Molarity and Normality (continued)

32. Determine the normality of a 0.2 M H₂SO₄ solution.

H2SO4→2H++SO42−\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}H2SO4→2H++SO42−

Step 1: Determine the number of equivalents:

Number of H+ions produced=2\text{Number of H}^+ \text{ions produced} = 2Number of H+ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.2 M×2=0.4 N\text{Normality} = 0.2 \, \text{M} \times 2 = 0.4 \, \text{N}Normality=0.2M×2=0.4N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

Additional Numerical Problems with Explanations

33. Calculate the molar mass of a gas if 2 g of the gas occupies 500 mL at STP.

Step 1: Use the ideal gas equation at STP:

At STP: 1 mole of gas occupies 22.4 L\text{At STP: 1 mole of gas occupies 22.4 L}At STP: 1 mole of gas occupies 22.4 L

Step 2: Determine the number of moles of the gas:

Volume of gas=500 mL=0.5 L\text{Volume of gas} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of gas=500mL=0.5L Moles of gas=0.5 L22.4 L/mol=0.0223 mol\text{Moles of gas} = \frac{0.5 \, \text{L}}{22.4 \, \text{L/mol}} = 0.0223 \, \text{mol}Moles of gas=22.4L/mol0.5L=0.0223mol

Step 3: Calculate the molar mass:

Molar mass=massmoles=2 g0.0223 mol=89.7 g/mol\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{2 \, \text{g}}{0.0223 \, \text{mol}} = 89.7 \, \text{g/mol}Molar mass=molesmass=0.0223mol2g=89.7g/mol

Explanation: Use the ideal gas law to find the number of moles, then divide the mass by the number of moles to find the molar mass.

34. Determine the mass of 0.5 moles of copper (Cu).

Atomic weight of Cu=63.5 g/mol\text{Atomic weight of Cu} = 63.5 \, \text{g/mol}Atomic weight of Cu=63.5g/mol Mass of Cu=0.5 mol×63.5 g/mol=31.75 g\text{Mass of Cu} = 0.5 \, \text{mol} \times 63.5 \, \text{g/mol} = 31.75 \, \text{g}Mass of Cu=0.5mol×63.5g/mol=31.75g

Explanation: Multiply the number of moles by the atomic weight to find the mass of the element.

35. Calculate the number of moles of HCl required to neutralize 0.5 moles of NaOH.

HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}HCl+NaOH→NaCl+H2O

Step 1: Use the stoichiometric ratio (1:1) to find moles of HCl:

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Moles of HCl=0.5 mol NaOH×1 mol HCl1 mol NaOH=0.5 mol HCl\text{Moles of HCl} = 0.5 \, \text{mol NaOH} \times \frac{1 \, \text{mol HCl}}{1 \, \text{mol NaOH}} = 0.5 \, \text{mol HCl}Moles of HCl=0.5mol NaOH×1mol NaOH1mol HCl=0.5mol HCl

Explanation: Use the stoichiometric ratio from the balanced equation to find the number of moles of HCl required.

Mole Concept (continued)

36. Calculate the volume of 0.25 moles of O₂ gas at STP.

At STP, 1 mole of gas occupies 22.4 L\text{At STP, 1 mole of gas occupies 22.4 L}At STP, 1 mole of gas occupies 22.4 L

Volume of O2=0.25 mol×22.4 L/mol=5.6 L\text{Volume of O}_2 = 0.25 \, \text{mol} \times 22.4 \, \text{L/mol} = 5.6 \, \text{L}Volume of O2=0.25mol×22.4L/mol=5.6L

Explanation: Multiply the number of moles by the molar volume of a gas at STP to find the volume.

Determination of Formula of Compound (continued)

37. A compound contains 36.5% sodium (Na), 25.4% sulfur (S), and 38.1% oxygen (O) by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Sodium: 36.5 g
Sulfur: 25.4 g
Oxygen: 38.1 g

Step 2: Convert mass to moles:

Moles of Na=36.5 g23 g/mol=1.587 mol\text{Moles of Na} = \frac{36.5 \, \text{g}}{23 \, \text{g/mol}} = 1.587 \, \text{mol}Moles of Na=23g/mol36.5g=1.587mol Moles of S=25.4 g32 g/mol=0.79375 mol\text{Moles of S} = \frac{25.4 \, \text{g}}{32 \, \text{g/mol}} = 0.79375 \, \text{mol}Moles of S=32g/mol25.4g=0.79375mol Moles of O=38.1 g16 g/mol=2.381 mol\text{Moles of O} = \frac{38.1 \, \text{g}}{16 \, \text{g/mol}} = 2.381 \, \text{mol}Moles of O=16g/mol38.1g=2.381mol

Step 3: Determine the simplest mole ratio:

Ratio of Na:S:O=1.587:0.79375:2.381≈2:1:3\text{Ratio of Na:S:O} = 1.587:0.79375:2.381 \approx 2:1:3Ratio of Na:S:O=1.587:0.79375:2.381≈2:1:3

Empirical formula: Na₂SO₃

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations (continued)

38. How many grams of NH₃ can be produced from 20 g of N₂ and excess H₂?

N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2+3H2→2NH3

Step 1: Determine moles of N₂:

Molecular weight of N2=2×14=28 g/mol\text{Molecular weight of N}_2 = 2 \times 14 = 28 \, \text{g/mol}Molecular weight of N2=2×14=28g/mol Moles of N2=20 g28 g/mol=0.714 mol\text{Moles of N}_2 = \frac{20 \, \text{g}}{28 \, \text{g/mol}} = 0.714 \, \text{mol}Moles of N2=28g/mol20g=0.714mol

Step 2: Use the stoichiometric ratio (1:2) to find moles of NH₃:

Moles of NH3=0.714 mol N2×2 mol NH31 mol N2=1.428 mol\text{Moles of NH}_3 = 0.714 \, \text{mol N}_2 \times \frac{2 \, \text{mol NH}_3}{1 \, \text{mol N}_2} = 1.428 \, \text{mol}Moles of NH3=0.714mol N2×1mol N22mol NH3=1.428mol

Step 3: Calculate the mass of NH₃:

Molecular weight of NH3=14+3×1=17 g/mol\text{Molecular weight of NH}_3 = 14 + 3 \times 1 = 17 \, \text{g/mol}Molecular weight of NH3=14+3×1=17g/mol Mass of NH3=1.428 mol×17 g/mol=24.276 g\text{Mass of NH}_3 = 1.428 \, \text{mol} \times 17 \, \text{g/mol} = 24.276 \, \text{g}Mass of NH3=1.428mol×17g/mol=24.276g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the product.

Concentration Terms (continued)

39. Calculate the molarity of a solution prepared by dissolving 5 g of Na₂SO₄ in 250 mL of solution.

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=moles of solutevolume of solution in liters=0.0352 mol0.25 L=0.141 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.0352 \, \text{mol}}{0.25 \, \text{L}} = 0.141 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.25L0.0352mol=0.141M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

Relation Between Molarity and Normality (continued)

40. Determine the normality of a 0.1 M H₃PO₄ solution.

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3\text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ions produced=3\text{Number of H}^+ \text{ions produced} = 3Number of H+ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.1 M×3=0.3 N\text{Normality} = 0.1 \, \text{M} \times 3 = 0.3 \, \text{N}Normality=0.1M×3=0.3N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

Additional Numerical Problems with Explanations

41. Calculate the mass of 0.75 moles of CaCO₃.

Molecular weight of CaCO3=40+12+3×16=100 g/mol\text{Molecular weight of CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol}Molecular weight of CaCO3=40+12+3×16=100g/mol Mass of CaCO3=0.75 mol×100 g/mol=75 g\text{Mass of CaCO}_3 = 0.75 \, \text{mol} \times 100 \, \text{g/mol} = 75 \, \text{g}Mass of CaCO3=0.75mol×100g/mol=75g

Explanation: Multiply the number of moles by the molecular weight to find the mass of the compound.

42. How many liters of hydrogen gas (H₂) at STP are produced when 5 g of zinc (Zn) reacts with excess hydrochloric acid (HCl)?

Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2Zn+2HCl→ZnCl2+H2

Step 1: Determine moles of Zn:

Molecular weight of Zn=65.38 g/mol\text{Molecular weight of Zn} = 65.38 \, \text{g/mol}Molecular weight of Zn=65.38g/mol Moles of Zn=5 g65.38 g/mol=0.0765 mol\text{Moles of Zn} = \frac{5 \, \text{g}}{65.38 \, \text{g/mol}} = 0.0765 \, \text{mol}Moles of Zn=65.38g/mol5g=0.0765mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of H₂:

Moles of H2=0.0765 mol Zn×1 mol H21 mol Zn=0.0765 mol\text{Moles of H}_2 = 0.0765 \, \text{mol Zn} \times \frac{1 \, \text{mol H}_2}{1 \, \text{mol Zn}} = 0.0765 \, \text{mol}Moles of H2=0.0765mol Zn×1mol Zn1mol H2=0.0765mol

Step 3: Calculate the volume of H₂ at STP:

Volume of H2=0.0765 mol×22.4 L/mol=1.7136 L\text{Volume of H}_2 = 0.0765 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.7136 \, \text{L}Volume of H2=0.0765mol×22.4L/mol=1.7136L

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the volume of the gas produced at STP.

43. Determine the molality of a solution prepared by dissolving 8 g of KOH in 200 g of water.

Molecular weight of KOH=39+16+1=56 g/mol\text{Molecular weight of KOH} = 39 + 16 + 1 = 56 \, \text{g/mol}Molecular weight of KOH=39+16+1=56g/mol Moles of KOH=8 g56 g/mol=0.1429 mol\text{Moles of KOH} = \frac{8 \, \text{g}}{56 \, \text{g/mol}} = 0.1429 \, \text{mol}Moles of KOH=56g/mol8g=0.1429mol

Mass of solvent (water)=200 g=0.2 kg\text{Mass of solvent (water)} = 200 \, \text{g} = 0.2 \, \text{kg}Mass of solvent (water)=200g=0.2kg Molality=moles of solutemass of solvent in kg=0.1429 mol0.2 kg=0.715 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1429 \, \text{mol}}{0.2 \, \text{kg}} = 0.715 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.2kg0.1429mol=0.715mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

44. Calculate the molarity of a solution prepared by dissolving 12 g of CH₃OH in enough water to make 200 mL of solution.

Molecular weight of CH3OH=12+3×1+16+1=32 g/mol\text{Molecular weight of CH}_3\text{OH} = 12 + 3 \times 1 + 16 + 1 = 32 \, \text{g/mol}Molecular weight of CH3OH=12+3×1+16+1=32g/mol Moles of CH3OH=12 g32 g/mol=0.375 mol\text{Moles of CH}_3\text{OH} = \frac{12 \, \text{g}}{32 \, \text{g/mol}} = 0.375 \, \text{mol}Moles of CH3OH=32g/mol12g=0.375mol

Volume of solution=200 mL=0.2 L\text{Volume of solution} = 200 \, \text{mL} = 0.2 \, \text{L}Volume of solution=200mL=0.2L Molarity=moles of solutevolume of solution in liters=0.375 mol0.2 L=1.875 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.375 \, \text{mol}}{0.2 \, \text{L}} = 1.875 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.2L0.375mol=1.875M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

45. How many grams of K₂SO₄ are needed to prepare 250 mL of a 0.5 M solution?

Molecular weight of K2SO4=2×39+32+4×16=174 g/mol\text{Molecular weight of K}_2\text{SO}_4 = 2 \times 39 + 32 + 4 \times 16 = 174 \, \text{g/mol}Molecular weight of K2SO4=2×39+32+4×16=174g/mol Moles of K2SO4=0.5 M×0.25 L=0.125 mol\text{Moles of K}_2\text{SO}_4 = 0.5 \, \text{M} \times 0.25 \, \text{L} = 0.125 \, \text{mol}Moles of K2SO4=0.5M×0.25L=0.125mol

Mass of K2SO4=0.125 mol×174 g/mol=21.75 g\text{Mass of K}_2\text{SO}_4 = 0.125 \, \text{mol} \times 174 \, \text{g/mol} = 21.75 \, \text{g}Mass of K2SO4=0.125mol×174g/mol=21.75g

Explanation: Use the molarity and volume to find the moles, then multiply by the molecular weight to find the mass of the solute needed.

46. How many liters of CO₂ at STP are produced when 10 g of CaCO₃ is decomposed?

CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2CaCO3→CaO+CO2

Step 1: Determine moles of CaCO₃:

Molecular weight of CaCO3=40+12+3×16=100 g/mol\text{Molecular weight of CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol}Molecular weight of CaCO3=40+12+3×16=100g/mol Moles of CaCO3=10 g100 g/mol=0.1 mol\text{Moles of CaCO}_3 = \frac{10 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{mol}Moles of CaCO3=100g/mol10g=0.1mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of CO₂:

Moles of CO2=0.1 mol CaCO3×1 mol CO21 mol CaCO3=0.1 mol\text{Moles of CO}_2 = 0.1 \, \text{mol CaCO}_3 \times \frac{1 \, \text{mol CO}_2}{1 \, \text{mol CaCO}_3} = 0.1 \, \text{mol}Moles of CO2=0.1mol CaCO3×1mol CaCO31mol CO2=0.1mol

Step 3: Calculate the volume of CO₂ at STP:

Volume of CO2=0.1 mol×22.4 L/mol=2.24 L\text{Volume of CO}_2 = 0.1 \, \text{mol} \times 22.4 \, \text{L/mol} = 2.24 \, \text{L}Volume of CO2=0.1mol×22.4L/mol=2.24L

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the volume of the gas produced at STP.

47. Determine the molarity of a solution containing 3.5 g of H₂SO₄ in 250 mL of solution.

Molecular weight of H2SO4=2×1+32+4×16=98 g/mol\text{Molecular weight of H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H2SO4=2×1+32+4×16=98g/mol Moles of H2SO4=3.5 g98 g/mol=0.0357 mol\text{Moles of H}_2\text{SO}_4 = \frac{3.5 \, \text{g}}{98 \, \text{g/mol}} = 0.0357 \, \text{mol}Moles of H2SO4=98g/mol3.5g=0.0357mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=moles of solutevolume of solution in liters=0.0357 mol0.25 L=0.143 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.0357 \, \text{mol}}{0.25 \, \text{L}} = 0.143 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.25L0.0357mol=0.143M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

48. How many grams of HCl are required to prepare 500 mL of a 0.2 M solution?

Molecular weight of HCl=1+35.5=36.5 g/mol\text{Molecular weight of HCl} = 1 + 35.5 = 36.5 \, \text{g/mol}Molecular weight of HCl=1+35.5=36.5g/mol Moles of HCl=0.2 M×0.5 L=0.1 mol\text{Moles of HCl} = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{mol}Moles of HCl=0.2M×0.5L=0.1mol

Mass of HCl=0.1 mol×36.5 g/mol=3.65 g\text{Mass of HCl} = 0.1 \, \text{mol} \times 36.5 \, \text{g/mol} = 3.65 \, \text{g}Mass of HCl=0.1mol×36.5g/mol=3.65g

Explanation: Use the molarity and volume to find the moles, then multiply by the molecular weight to find the mass of the solute needed.

49. Calculate the molality of a solution prepared by dissolving 10 g of NaCl in 100 g of water.

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Moles of NaCl=10 g58.5 g/mol=0.171 mol\text{Moles of NaCl} = \frac{10 \, \text{g}}{58.5 \, \text{g/mol}} = 0.171 \, \text{mol}Moles of NaCl=58.5g/mol10g=0.171mol

Mass of solvent (water)=100 g=0.1 kg\text{Mass of solvent (water)} = 100 \, \text{g} = 0.1 \, \text{kg}Mass of solvent (water)=100g=0.1kg Molality=moles of solutemass of solvent in kg=0.171 mol0.1 kg=1.71 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.171 \, \text{mol}}{0.1 \, \text{kg}} = 1.71 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.1kg0.171mol=1.71mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

50. Determine the normality of a 0.5 M H₂C₂O₄ solution.

H2C2O4→2H++C2O42−\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{H}^+ + \text{C}_2\text{O}_4^{2-}H2C2O4→2H++C2O42−

Step 1: Determine the number of equivalents:

Number of H+ions produced=2\text{Number of H}^+ \text{ions produced} = 2Number of H+ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×2=1 N\text{Normality} = 0.5 \, \text{M} \times 2 = 1 \, \text{N}Normality=0.5M×2=1N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

CONTINUE

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Mole Concept (continued)

51. How many atoms are there in 2 moles of sodium (Na)?

Number of atoms=2 mol×6.022×1023 atoms/mol=1.2044×1024 atoms\text{Number of atoms} = 2 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 1.2044 \times 10^{24} \, \text{atoms}Number of atoms=2mol×6.022×1023atoms/mol=1.2044×1024atoms

Explanation: Multiply the number of moles by Avogadro's number to find the total number of atoms.

Determination of Formula of Compound (continued)

52. A compound contains 40% carbon (C), 6.7% hydrogen (H), and 53.3% oxygen (O) by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 40 g
Hydrogen: 6.7 g
Oxygen: 53.3 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=3.33:6.7:3.33≈1:2:1\text{Ratio of C:H:O} = 3.33:6.7:3.33 \approx 1:2:1Ratio of C:H:O=3.33:6.7:3.33≈1:2:1

Empirical formula: CH₂O

Explanation: Convert percentages to moles and find the simplest whole number ratio.

Stoichiometric Calculations (continued)

53. How many grams of H₂O are produced when 2 g of H₂ reacts with excess O₂?

2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2+O2→2H2O

Step 1: Determine moles of H₂:

Molecular weight of H2=2 g/mol\text{Molecular weight of H}_2 = 2 \, \text{g/mol}Molecular weight of H2=2g/mol Moles of H2=2 g2 g/mol=1 mol\text{Moles of H}_2 = \frac{2 \, \text{g}}{2 \, \text{g/mol}} = 1 \, \text{mol}Moles of H2=2g/mol2g=1mol

Step 2: Use the stoichiometric ratio (2:2) to find moles of H₂O:

Moles of H2O=1 mol H2×2 mol H2O2 mol H2=1 mol\text{Moles of H}_2\text{O} = 1 \, \text{mol H}_2 \times \frac{2 \, \text{mol H}_2\text{O}}{2 \, \text{mol H}_2} = 1 \, \text{mol}Moles of H2O=1mol H2×2mol H22mol H2O=1mol

Step 3: Calculate the mass of H₂O:

Molecular weight of H2O=2×1+16=18 g/mol\text{Molecular weight of H}_2\text{O} = 2 \times 1 + 16 = 18 \, \text{g/mol}Molecular weight of H2O=2×1+16=18g/mol Mass of H2O=1 mol×18 g/mol=18 g\text{Mass of H}_2\text{O} = 1 \, \text{mol} \times 18 \, \text{g/mol} = 18 \, \text{g}Mass of H2O=1mol×18g/mol=18g

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the mass of the product.

Concentration Terms (continued)

54. Calculate the molarity of a solution prepared by dissolving 15 g of NaCl in 500 mL of solution.

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Moles of NaCl=15 g58.5 g/mol=0.256 mol\text{Moles of NaCl} = \frac{15 \, \text{g}}{58.5 \, \text{g/mol}} = 0.256 \, \text{mol}Moles of NaCl=58.5g/mol15g=0.256mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=moles of solutevolume of solution in liters=0.256 mol0.5 L=0.512 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.256 \, \text{mol}}{0.5 \, \text{L}} = 0.512 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.5L0.256mol=0.512M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

Relation Between Molarity and Normality (continued)

55. Determine the normality of a 0.25 M H₃PO₄ solution.

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3\text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ions produced=3\text{Number of H}^+ \text{ions produced} = 3Number of H+ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.25 M×3=0.75 N\text{Normality} = 0.25 \, \text{M} \times 3 = 0.75 \, \text{N}Normality=0.25M×3=0.75N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

Additional Numerical Problems with Explanations (continued)

56. How many moles of NaOH are required to neutralize 100 mL of 1 M H₂SO₄?

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Determine moles of H₂SO₄:

Moles of H2SO4=1 M×0.1 L=0.1 mol\text{Moles of H}_2\text{SO}_4 = 1 \, \text{M} \times 0.1 \, \text{L} = 0.1 \, \text{mol}Moles of H2SO4=1M×0.1L=0.1mol

Step 2: Use the stoichiometric ratio (1:2) to find moles of NaOH:

Moles of NaOH=0.1 mol H2SO4×2 mol NaOH1 mol H2SO4=0.2 mol\text{Moles of NaOH} = 0.1 \, \text{mol H}_2\text{SO}_4 \times \frac{2 \, \text{mol NaOH}}{1 \, \text{mol H}_2\text{SO}_4} = 0.2 \, \text{mol}Moles of NaOH=0.1mol H2SO4×1mol H2SO42mol NaOH=0.2mol

Explanation: Use the stoichiometric coefficients from the balanced equation to find the moles of the reactant required for neutralization.

57. Calculate the mass of 0.2 moles of K₂CO₃.

Molecular weight of K2CO3=2×39+12+3×16=138 g/mol\text{Molecular weight of K}_2\text{CO}_3 = 2 \times 39 + 12 + 3 \times 16 = 138 \, \text{g/mol}Molecular weight of K2CO3=2×39+12+3×16=138g/mol Mass of K2CO3=0.2 mol×138 g/mol=27.6 g\text{Mass of K}_2\text{CO}_3 = 0.2 \, \text{mol} \times 138 \, \text{g/mol} = 27.6 \, \text{g}Mass of K2CO3=0.2mol×138g/mol=27.6g

Explanation: Multiply the number of moles by the molecular weight to find the mass of the compound.

58. How many liters of CH₄ at STP are produced when 5 g of carbon reacts with excess hydrogen?

C+2H2→CH4\text{C} + 2\text{H}_2 \rightarrow \text{CH}_4C+2H2→CH4

Step 1: Determine moles of carbon:

Molecular weight of C=12 g/mol\text{Molecular weight of C} = 12 \, \text{g/mol}Molecular weight of C=12g/mol Moles of C=5 g12 g/mol=0.417 mol\text{Moles of C} = \frac{5 \, \text{g}}{12 \, \text{g/mol}} = 0.417 \, \text{mol}Moles of C=12g/mol5g=0.417mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of CH₄:

Moles of CH4=0.417 mol C×1 mol CH41 mol C=0.417 mol\text{Moles of CH}_4 = 0.417 \, \text{mol C} \times \frac{1 \, \text{mol CH}_4}{1 \, \text{mol C}} = 0.417 \, \text{mol}Moles of CH4=0.417mol C×1mol C1mol CH4=0.417mol

Step 3: Calculate the volume of CH₄ at STP:

Volume of CH4=0.417 mol×22.4 L/mol=9.34 L\text{Volume of CH}_4 = 0.417 \, \text{mol} \times 22.4 \, \text{L/mol} = 9.34 \, \text{L}Volume of CH4=0.417mol×22.4L/mol=9.34L

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the volume of the gas produced at STP.

59. Determine the molarity of a solution containing 4 g of NaOH in 250 mL of solution.

Molecular weight of NaOH=23+16+1=40 g/mol\text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}Molecular weight of NaOH=23+16+1=40g/mol Moles of NaOH=4 g40 g/mol=0.1 mol\text{Moles of NaOH} = \frac{4 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol}Moles of NaOH=40g/mol4g=0.1mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=moles of solutevolume of solution in liters=0.1 mol0.25 L=0.4 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.1 \, \text{mol}}{0.25 \, \text{L}} = 0.4 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.25L0.1mol=0.4M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

60. Calculate the molality of a solution prepared by dissolving 5 g of glucose (C₆H₁₂O₆) in 100 g of water.

Mass of solvent (water)=100 g=0.1 kg\text{Mass of solvent (water)} = 100 \, \text{g} = 0.1 \, \text{kg}Mass of solvent (water)=100g=0.1kg Molality=moles of solutemass of solvent in kg=0.0278 mol0.1 kg=0.278 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0278 \, \text{mol}}{0.1 \, \text{kg}} = 0.278 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.1kg0.0278mol=0.278mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

61. How many grams of H₂SO₄ are required to prepare 1 L of a 2 M solution?

Molecular weight of H2SO4=2×1+32+4×16=98 g/mol\text{Molecular weight of H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H2SO4=2×1+32+4×16=98g/mol Moles of H2SO4=2 M×1 L=2 mol\text{Moles of H}_2\text{SO}_4 = 2 \, \text{M} \times 1 \, \text{L} = 2 \, \text{mol}Moles of H2SO4=2M×1L=2mol

Mass of H2SO4=2 mol×98 g/mol=196 g\text{Mass of H}_2\text{SO}_4 = 2 \, \text{mol} \times 98 \, \text{g/mol} = 196 \, \text{g}Mass of H2SO4=2mol×98g/mol=196g

Explanation: Use the molarity and volume to find the moles, then multiply by the molecular weight to find the mass of the solute needed.

62. Calculate the normality of a solution prepared by dissolving 49 g of H₂SO₄ in 500 mL of solution.

Molecular weight of H2SO4=2×1+32+4×16=98 g/mol\text{Molecular weight of H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H2SO4=2×1+32+4×16=98g/mol Moles of H2SO4=49 g98 g/mol=0.5 mol\text{Moles of H}_2\text{SO}_4 = \frac{49 \, \text{g}}{98 \, \text{g/mol}} = 0.5 \, \text{mol}Moles of H2SO4=98g/mol49g=0.5mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=0.5 mol0.5 L=1 M\text{Molarity} = \frac{0.5 \, \text{mol}}{0.5 \, \text{L}} = 1 \, \text{M}Molarity=0.5L0.5mol=1M

Number of equivalents=2\text{Number of equivalents} = 2Number of equivalents=2 Normality=1 M×2=2 N\text{Normality} = 1 \, \text{M} \times 2 = 2 \, \text{N}Normality=1M×2=2N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

63. Determine the volume of 0.5 M HCl needed to completely neutralize 250 mL of 0.2 M NaOH.

HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}HCl+NaOH→NaCl+H2O

Step 1: Determine moles of NaOH:

Moles of NaOH=0.2 M×0.25 L=0.05 mol\text{Moles of NaOH} = 0.2 \, \text{M} \times 0.25 \, \text{L} = 0.05 \, \text{mol}Moles of NaOH=0.2M×0.25L=0.05mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of HCl:

Moles of HCl=0.05 mol\text{Moles of HCl} = 0.05 \, \text{mol}Moles of HCl=0.05mol

Step 3: Calculate the volume of HCl:

Volume of HCl=moles of solutemolarity=0.05 mol0.5 M=0.1 L=100 mL\text{Volume of HCl} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.05 \, \text{mol}}{0.5 \, \text{M}} = 0.1 \, \text{L} = 100 \, \text{mL}Volume of HCl=molaritymoles of solute=0.5M0.05mol=0.1L=100mL

Explanation: Use the stoichiometric coefficients from the balanced equation and the molarity to find the volume of the reactant needed for neutralization.

64. How many grams of CaCl₂ are required to prepare 250 mL of a 0.4 M solution?

Molecular weight of CaCl2=40+2×35.5=111 g/mol\text{Molecular weight of CaCl}_2 = 40 + 2 \times 35.5 = 111 \, \text{g/mol}Molecular weight of CaCl2=40+2×35.5=111g/mol Moles of CaCl2=0.4 M×0.25 L=0.1 mol\text{Moles of CaCl}_2 = 0.4 \, \text{M} \times 0.25 \, \text{L} = 0.1 \, \text{mol}Moles of CaCl2=0.4M×0.25L=0.1mol

Mass of CaCl2=0.1 mol×111 g/mol=11.1 g\text{Mass of CaCl}_2 = 0.1 \, \text{mol} \times 111 \, \text{g/mol} = 11.1 \, \text{g}Mass of CaCl2=0.1mol×111g/mol=11.1g

Explanation: Use the molarity and volume to find the moles, then multiply by the molecular weight to find the mass of the solute needed.

65. Calculate the molality of a solution prepared by dissolving 20 g of KOH in 150 g of water.

Molecular weight of KOH=39+16+1=56 g/mol\text{Molecular weight of KOH} = 39 + 16 + 1 = 56 \, \text{g/mol}Molecular weight of KOH=39+16+1=56g/mol Moles of KOH=20 g56 g/mol=0.357 mol\text{Moles of KOH} = \frac{20 \, \text{g}}{56 \, \text{g/mol}} = 0.357 \, \text{mol}Moles of KOH=56g/mol20g=0.357mol

Mass of solvent (water)=150 g=0.15 kg\text{Mass of solvent (water)} = 150 \, \text{g} = 0.15 \, \text{kg}Mass of solvent (water)=150g=0.15kg Molality=moles of solutemass of solvent in kg=0.357 mol0.15 kg=2.38 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.357 \, \text{mol}}{0.15 \, \text{kg}} = 2.38 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.15kg0.357mol=2.38mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

66. How many liters of H₂ at STP are produced when 5 g of Zn reacts with excess HCl?

Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2Zn+2HCl→ZnCl2+H2

Step 1: Determine moles of Zn:

Molecular weight of Zn=65 g/mol\text{Molecular weight of Zn} = 65 \, \text{g/mol}Molecular weight of Zn=65g/mol Moles of Zn=5 g65 g/mol=0.0769 mol\text{Moles of Zn} = \frac{5 \, \text{g}}{65 \, \text{g/mol}} = 0.0769 \, \text{mol}Moles of Zn=65g/mol5g=0.0769mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of H₂:

Moles of H2=0.0769 mol Zn×1 mol H21 mol Zn=0.0769 mol\text{Moles of H}_2 = 0.0769 \, \text{mol Zn} \times \frac{1 \, \text{mol H}_2}{1 \, \text{mol Zn}} = 0.0769 \, \text{mol}Moles of H2=0.0769mol Zn×1mol Zn1mol H2=0.0769mol

Step 3: Calculate the volume of H₂ at STP:

Volume of H2=0.0769 mol×22.4 L/mol=1.72 L\text{Volume of H}_2 = 0.0769 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.72 \, \text{L}Volume of H2=0.0769mol×22.4L/mol=1.72L

Explanation: Use the mole concept and stoichiometric coefficients from the balanced equation to find the volume of the gas produced at STP.

67. Determine the molarity of a solution containing 10 g of KNO₃ in 200 mL of solution.

Molecular weight of KNO3=39+14+3×16=101 g/mol\text{Molecular weight of KNO}_3 = 39 + 14 + 3 \times 16 = 101 \, \text{g/mol}Molecular weight of KNO3=39+14+3×16=101g/mol Moles of KNO3=10 g101 g/mol=0.099 mol\text{Moles of KNO}_3 = \frac{10 \, \text{g}}{101 \, \text{g/mol}} = 0.099 \, \text{mol}Moles of KNO3=101g/mol10g=0.099mol

Volume of solution=200 mL=0.2 L\text{Volume of solution} = 200 \, \text{mL} = 0.2 \, \text{L}Volume of solution=200mL=0.2L Molarity=moles of solutevolume of solution in liters=0.099 mol0.2 L=0.495 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.099 \, \text{mol}}{0.2 \, \text{L}} = 0.495 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.2L0.099mol=0.495M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

68. Calculate the normality of a solution prepared by dissolving 50 g of NaOH in 2 L of solution.

Molecular weight of NaOH=23+16+1=40 g/mol\text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}Molecular weight of NaOH=23+16+1=40g/mol Moles of NaOH=50 g40 g/mol=1.25 mol\text{Moles of NaOH} = \frac{50 \, \text{g}}{40 \, \text{g/mol}} = 1.25 \, \text{mol}Moles of NaOH=40g/mol50g=1.25mol

Volume of solution=2 L\text{Volume of solution} = 2 \, \text{L}Volume of solution=2L Molarity=1.25 mol2 L=0.625 M\text{Molarity} = \frac{1.25 \, \text{mol}}{2 \, \text{L}} = 0.625 \, \text{M}Molarity=2L1.25mol=0.625M

Number of equivalents=1\text{Number of equivalents} = 1Number of equivalents=1 Normality=0.625 M×1=0.625 N\text{Normality} = 0.625 \, \text{M} \times 1 = 0.625 \, \text{N}Normality=0.625M×1=0.625N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

69. Determine the volume of 0.1 M NaOH needed to neutralize 50 mL of 0.05 M HCl.

NaOH+HCl→NaCl+H2O\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}NaOH+HCl→NaCl+H2O

Step 1: Determine moles of HCl:

Moles of HCl=0.05 M×0.05 L=0.0025 mol\text{Moles of HCl} = 0.05 \, \text{M} \times 0.05 \, \text{L} = 0.0025 \, \text{mol}Moles of HCl=0.05M×0.05L=0.0025mol

Step 2: Use the stoichiometric ratio (1:1) to find moles of NaOH:

Moles of NaOH=0.0025 mol\text{Moles of NaOH} = 0.0025 \, \text{mol}Moles of NaOH=0.0025mol

Step 3: Calculate the volume of NaOH:

Volume of NaOH=moles of solutemolarity=0.0025 mol0.1 M=0.025 L=25 mL\text{Volume of NaOH} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.0025 \, \text{mol}}{0.1 \, \text{M}} = 0.025 \, \text{L} = 25 \, \text{mL}Volume of NaOH=molaritymoles of solute=0.1M0.0025mol=0.025L=25mL

Explanation: Use the stoichiometric coefficients from the balanced equation and the molarity to find the volume of the reactant needed for neutralization.

70. Calculate the molality of a solution prepared by dissolving 10 g of NaCl in 500 g of water.

Mass of solvent (water)=500 g=0.5 kg\text{Mass of solvent (water)} = 500 \, \text{g} = 0.5 \, \text{kg}Mass of solvent (water)=500g=0.5kg Molality=moles of solutemass of solvent in kg=0.171 mol0.5 kg=0.342 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.171 \, \text{mol}}{0.5 \, \text{kg}} = 0.342 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.5kg0.171mol=0.342mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

This concludes the numerical problems section. Each problem is designed to enhance your understanding of the basic concepts of chemistry, particularly focusing on the mole concept, formula determination, stoichiometric calculations, and concentration terms.

CONTINUE

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Mole Concept (continued)

71. Calculate the number of molecules in 5 moles of CO₂.

Number of molecules=5 mol×6.022×1023 molecules/mol=3.011×1024 molecules\text{Number of molecules} = 5 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 3.011 \times 10^{24} \, \text{molecules}Number of molecules=5mol×6.022×1023molecules/mol=3.011×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to determine the number of molecules.

Determination of Formula of Compound (continued)

72. A compound contains 30% carbon, 10% hydrogen, and 60% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 30 g
Hydrogen: 10 g
Oxygen: 60 g

Step 2: Convert mass to moles:

Moles of C=30 g12 g/mol=2.5 mol\text{Moles of C} = \frac{30 \, \text{g}}{12 \, \text{g/mol}} = 2.5 \, \text{mol}Moles of C=12g/mol30g=2.5mol Moles of H=10 g1 g/mol=10 mol\text{Moles of H} = \frac{10 \, \text{g}}{1 \, \text{g/mol}} = 10 \, \text{mol}Moles of H=1g/mol10g=10mol Moles of O=60 g16 g/mol=3.75 mol\text{Moles of O} = \frac{60 \, \text{g}}{16 \, \text{g/mol}} = 3.75 \, \text{mol}Moles of O=16g/mol60g=3.75mol

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=2.5:10:3.75≈2:8:6\text{Ratio of C:H:O} = 2.5:10:3.75 \approx 2:8:6Ratio of C:H:O=2.5:10:3.75≈2:8:6

Step 4: Simplify to whole numbers:

Empirical formula=C2H8O6\text{Empirical formula} = \text{C}_2\text{H}_8\text{O}_6Empirical formula=C2H8O6

Explanation: Convert percentages to moles and simplify to the smallest whole number ratio.

Stoichiometric Calculations (continued)

73. How many moles of O₂ are required to react completely with 4 moles of C₃H₈?

C3H8+5O2→3CO2+4H2O\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}C3H8+5O2→3CO2+4H2O

Step 1: Use the stoichiometric ratio (1:5):

Moles of O2=4 mol C3H8×5 mol O21 mol C3H8=20 mol\text{Moles of O}_2 = 4 \, \text{mol C}_3\text{H}_8 \times \frac{5 \, \text{mol O}_2}{1 \, \text{mol C}_3\text{H}_8} = 20 \, \text{mol}Moles of O2=4mol C3H8×1mol C3H85mol O2=20mol

Explanation: Multiply the moles of the reactant by the ratio of moles required for the other reactant.

Concentration Terms (continued)

74. Calculate the concentration in terms of molality of a solution containing 25 g of Na₂SO₄ in 200 g of water.

Mass of solvent (water)=200 g=0.2 kg\text{Mass of solvent (water)} = 200 \, \text{g} = 0.2 \, \text{kg}Mass of solvent (water)=200g=0.2kg Molality=moles of solutemass of solvent in kg=0.176 mol0.2 kg=0.88 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.176 \, \text{mol}}{0.2 \, \text{kg}} = 0.88 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.2kg0.176mol=0.88mol/kg

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent.

Relation Between Molarity and Normality (continued)

75. Determine the normality of a 0.5 M H₂SO₄ solution.

H2SO4→2H++SO42−\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}H2SO4→2H++SO42−

Step 1: Determine the number of equivalents:

Number of H+ions produced=2\text{Number of H}^+ \text{ions produced} = 2Number of H+ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×2=1 N\text{Normality} = 0.5 \, \text{M} \times 2 = 1 \, \text{N}Normality=0.5M×2=1N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

Additional Numerical Problems with Explanations (continued)

76. Calculate the mass of 3 moles of NaHCO₃.

Molecular weight of NaHCO3=23+1+12+3×16=84 g/mol\text{Molecular weight of NaHCO}_3 = 23 + 1 + 12 + 3 \times 16 = 84 \, \text{g/mol}Molecular weight of NaHCO3=23+1+12+3×16=84g/mol Mass of NaHCO3=3 mol×84 g/mol=252 g\text{Mass of NaHCO}_3 = 3 \, \text{mol} \times 84 \, \text{g/mol} = 252 \, \text{g}Mass of NaHCO3=3mol×84g/mol=252g

Explanation: Multiply the number of moles by the molecular weight to find the mass of the compound.

77. Determine the volume of 0.2 M HNO₃ needed to react with 0.1 mol of NaOH.

HNO3+NaOH→NaNO3+H2O\text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}HNO3+NaOH→NaNO3+H2O

Step 1: Use the stoichiometric ratio (1:1) to find moles of HNO₃:

Moles of HNO3=0.1 mol NaOH×1 mol HNO31 mol NaOH=0.1 mol\text{Moles of HNO}_3 = 0.1 \, \text{mol NaOH} \times \frac{1 \, \text{mol HNO}_3}{1 \, \text{mol NaOH}} = 0.1 \, \text{mol}Moles of HNO3=0.1mol NaOH×1mol NaOH1mol HNO3=0.1mol

Step 2: Calculate the volume of HNO₃:

Volume of HNO3=moles of solutemolarity=0.1 mol0.2 M=0.5 L=500 mL\text{Volume of HNO}_3 = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.1 \, \text{mol}}{0.2 \, \text{M}} = 0.5 \, \text{L} = 500 \, \text{mL}Volume of HNO3=molaritymoles of solute=0.2M0.1mol=0.5L=500mL

Explanation: Use the stoichiometric coefficients from the balanced equation and the molarity to find the volume of the reactant needed for reaction.

78. Calculate the mass of 4 moles of K₂SO₄.

Molecular weight of K2SO4=2×39+32+4×16=174 g/mol\text{Molecular weight of K}_2\text{SO}_4 = 2 \times 39 + 32 + 4 \times 16 = 174 \, \text{g/mol}Molecular weight of K2SO4=2×39+32+4×16=174g/mol Mass of K2SO4=4 mol×174 g/mol=696 g\text{Mass of K}_2\text{SO}_4 = 4 \, \text{mol} \times 174 \, \text{g/mol} = 696 \, \text{g}Mass of K2SO4=4mol×174g/mol=696g

Explanation: Multiply the number of moles by the molecular weight to find the mass of the compound.

79. How many liters of O₂ are required to react completely with 5 L of CH₄ at STP?

CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}CH4+2O2→CO2+2H2O

Step 1: Use the stoichiometric ratio (1:2) to find the volume of O₂:

Volume of O2=5 L CH4×2 L O21 L CH4=10 L\text{Volume of O}_2 = 5 \, \text{L CH}_4 \times \frac{2 \, \text{L O}_2}{1 \, \text{L CH}_4} = 10 \, \text{L}Volume of O2=5L CH4×1L CH42L O2=10L

Explanation: Use the stoichiometric coefficients from the balanced equation to find the volume of the reactant needed.

80. Determine the molarity of a solution prepared by dissolving 40 g of NaCl in 250 mL of solution.

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Moles of NaCl=40 g58.5 g/mol=0.683 mol\text{Moles of NaCl} = \frac{40 \, \text{g}}{58.5 \, \text{g/mol}} = 0.683 \, \text{mol}Moles of NaCl=58.5g/mol40g=0.683mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=moles of solutevolume of solution in liters=0.683 mol0.25 L=2.732 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.683 \, \text{mol}}{0.25 \, \text{L}} = 2.732 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.25L0.683mol=2.732M

Explanation: Molarity is defined as the number of moles of solute per liter of solution.

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Mole Concept (continued)

81. How many moles are present in 50 g of Na₂CO₃?

Molecular weight of Na2CO3=2×23+12+3×16=106 g/mol\text{Molecular weight of Na}_2\text{CO}_3 = 2 \times 23 + 12 + 3 \times 16 = 106 \, \text{g/mol}Molecular weight of Na2CO3=2×23+12+3×16=106g/mol Moles of Na2CO3=50 g106 g/mol=0.472 mol\text{Moles of Na}_2\text{CO}_3 = \frac{50 \, \text{g}}{106 \, \text{g/mol}} = 0.472 \, \text{mol}Moles of Na2CO3=106g/mol50g=0.472mol

Explanation: Divide the mass of the compound by its molecular weight to find the number of moles.

82. Determine the number of atoms in 2 moles of Fe.

Explanation: Multiply the number of moles by Avogadro's number to determine the number of atoms.

Determination of Formula of Compound (continued)

83. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 40 g
Hydrogen: 6.7 g
Oxygen: 53.3 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=3.33:6.7:3.33≈1:2:1\text{Ratio of C:H:O} = 3.33:6.7:3.33 \approx 1:2:1Ratio of C:H:O=3.33:6.7:3.33≈1:2:1

Step 4: Empirical formula:

Empirical formula=CH2O\text{Empirical formula} = \text{CH}_2\text{O}Empirical formula=CH2O

Explanation: Convert percentages to moles, then simplify the mole ratio to get the empirical formula.

Stoichiometric Calculations (continued)

84. How many grams of CaCO₃ are needed to react completely with 0.5 mol of HCl?

CaCO3+2HCl→CaCl2+CO2+H2O\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}CaCO3+2HCl→CaCl2+CO2+H2O

Step 1: Determine moles of CaCO₃ required:

Moles of CaCO3=0.5 mol HCl×1 mol CaCO32 mol HCl=0.25 mol\text{Moles of CaCO}_3 = 0.5 \, \text{mol HCl} \times \frac{1 \, \text{mol CaCO}_3}{2 \, \text{mol HCl}} = 0.25 \, \text{mol}Moles of CaCO3=0.5mol HCl×2mol HCl1mol CaCO3=0.25mol

Step 2: Calculate the mass of CaCO₃:

Molecular weight of CaCO3=40+12+3×16=100 g/mol\text{Molecular weight of CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol}Molecular weight of CaCO3=40+12+3×16=100g/mol Mass of CaCO3=0.25 mol×100 g/mol=25 g\text{Mass of CaCO}_3 = 0.25 \, \text{mol} \times 100 \, \text{g/mol} = 25 \, \text{g}Mass of CaCO3=0.25mol×100g/mol=25g

Explanation: Use stoichiometric ratios to determine the moles of the reactant needed, then find the mass.

Concentration Terms (continued)

85. Calculate the molarity of a solution prepared by dissolving 15 g of KCl in 500 mL of solution.

Molecular weight of KCl=39+35.5=74.5 g/mol\text{Molecular weight of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol}Molecular weight of KCl=39+35.5=74.5g/mol Moles of KCl=15 g74.5 g/mol=0.201 mol\text{Moles of KCl} = \frac{15 \, \text{g}}{74.5 \, \text{g/mol}} = 0.201 \, \text{mol}Moles of KCl=74.5g/mol15g=0.201mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=moles of solutevolume of solution in liters=0.201 mol0.5 L=0.402 M\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.201 \, \text{mol}}{0.5 \, \text{L}} = 0.402 \, \text{M}Molarity=volume of solution in litersmoles of solute=0.5L0.201mol=0.402M

Explanation: Molarity is calculated by dividing the number of moles by the volume of solution in liters.

Relation Between Molarity and Normality (continued)

86. Determine the normality of a 0.1 M H₃PO₄ solution.

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3\text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ions produced=3\text{Number of H}^+ \text{ions produced} = 3Number of H+ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.1 M×3=0.3 N\text{Normality} = 0.1 \, \text{M} \times 3 = 0.3 \, \text{N}Normality=0.1M×3=0.3N

Explanation: Normality is calculated by multiplying the molarity by the number of equivalents per mole of solute.

Additional Numerical Problems with Explanations (continued)

87. Calculate the volume of 0.5 M NaOH needed to neutralize 100 mL of 0.25 M H₂SO₄.

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Determine moles of H₂SO₄:

Moles of H2SO4=0.25 M×0.1 L=0.025 mol\text{Moles of H}_2\text{SO}_4 = 0.25 \, \text{M} \times 0.1 \, \text{L} = 0.025 \, \text{mol}Moles of H2SO4=0.25M×0.1L=0.025mol

Step 2: Use the stoichiometric ratio (1:2) to find moles of NaOH:

Moles of NaOH=0.025 mol H2SO4×2 mol NaOH1 mol H2SO4=0.05 mol\text{Moles of NaOH} = 0.025 \, \text{mol H}_2\text{SO}_4 \times \frac{2 \, \text{mol NaOH}}{1 \, \text{mol H}_2\text{SO}_4} = 0.05 \, \text{mol}Moles of NaOH=0.025mol H2SO4×1mol H2SO42mol NaOH=0.05mol

Step 3: Calculate the volume of NaOH:

Volume of NaOH=moles of solutemolarity=0.05 mol0.5 M=0.1 L=100 mL\text{Volume of NaOH} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.05 \, \text{mol}}{0.5 \, \text{M}} = 0.1 \, \text{L} = 100 \, \text{mL}Volume of NaOH=molaritymoles of solute=0.5M0.05mol=0.1L=100mL

Explanation: Use the stoichiometric coefficients from the balanced equation and the molarity to find the volume of the reactant needed for neutralization.

88. How many grams of Na₂SO₄ are required to prepare 500 mL of a 0.1 M solution?

Molecular weight of Na2SO4=2×23+32+4×16=142 g/mol\text{Molecular weight of Na}_2\text{SO}_4 = 2 \times 23 + 32 + 4 \times 16 = 142 \, \text{g/mol}Molecular weight of Na2SO4=2×23+32+4×16=142g/mol Moles of Na2SO4=0.1 M×0.5 L=0.05 mol\text{Moles of Na}_2\text{SO}_4 = 0.1 \, \text{M} \times 0.5 \, \text{L} = 0.05 \, \text{mol}Moles of Na2SO4=0.1M×0.5L=0.05mol Mass of Na2SO4=0.05 mol×142 g/mol=7.1 g\text{Mass of Na}_2\text{SO}_4 = 0.05 \, \text{mol} \times 142 \, \text{g/mol} = 7.1 \, \text{g}Mass of Na2SO4=0.05mol×142g/mol=7.1g

Explanation: Use the molarity and volume to find the moles, then multiply by the molecular weight to find the mass of the solute needed.

89. Determine the normality of a solution prepared by dissolving 10 g of H₂SO₄ in 250 mL of solution.

Molecular weight of H2SO4=2×1+32+4×16=98 g/mol\text{Molecular weight of H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H2SO4=2×1+32+4×16=98g/mol Moles of H2SO4=10 g98 g/mol=0.102 mol\text{Moles of H}_2\text{SO}_4 = \frac{10 \, \text{g}}{98 \, \text{g/mol}} = 0.102 \, \text{mol}Moles of H2SO4=98g/mol10g=0.102mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=0.102 mol0.25 L=0.408 M\text{Molarity} = \frac{0.102 \, \text{mol}}{0.25 \, \text{L}} = 0.408 \, \text{M}Molarity=0.25L0.102mol=0.408M

Number of equivalents=2\text{Number of equivalents} = 2Number of equivalents=2 Normality=0.408 M×2=0.816 N\text{Normality} = 0.408 \, \text{M} \times 2 = 0.816 \, \text{N}Normality=0.408M×2=0.816N

Explanation: Calculate molarity first, then use the number of equivalents to find the normality.

90. Calculate the molality of a solution prepared by dissolving 20 g of glucose (C₆H₁₂O₆) in 400 g of water.

Mass of solvent (water)=400 g=0.4 kg\text{Mass of solvent (water)} = 400 \, \text{g} = 0.4 \, \text{kg}Mass of solvent (water)=400g=0.4kg Molality=moles of solutemass of solvent in kg=0.111 mol0.4 kg=0.278 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.111 \, \text{mol}}{0.4 \, \text{kg}} = 0.278 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.4kg0.111mol=0.278mol/kg

Explanation: Molality is calculated by dividing the number of moles of solute by the mass of solvent in kilograms.

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Mole Concept (continued)

91. Calculate the number of atoms in 0.5 moles of Mg.

Number of atoms=0.5 mol×6.022×1023 atoms/mol=3.011×1023 atoms\text{Number of atoms} = 0.5 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 3.011 \times 10^{23} \, \text{atoms}Number of atoms=0.5mol×6.022×1023atoms/mol=3.011×1023atoms

Explanation: Multiply the number of moles by Avogadro's number to find the number of atoms.

92. How many moles of Fe are present in 15 g of Fe?

Molecular weight of Fe=56 g/mol\text{Molecular weight of Fe} = 56 \, \text{g/mol}Molecular weight of Fe=56g/mol Moles of Fe=15 g56 g/mol=0.268 mol\text{Moles of Fe} = \frac{15 \, \text{g}}{56 \, \text{g/mol}} = 0.268 \, \text{mol}Moles of Fe=56g/mol15g=0.268mol

Explanation: Divide the mass of Fe by its molecular weight to find the number of moles.

Determination of Formula of Compound (continued)

93. A compound contains 52.2% carbon, 34.6% oxygen, and 13.2% hydrogen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 52.2 g
Hydrogen: 13.2 g
Oxygen: 34.6 g

Step 2: Convert mass to moles:

Moles of C=52.2 g12 g/mol=4.35 mol\text{Moles of C} = \frac{52.2 \, \text{g}}{12 \, \text{g/mol}} = 4.35 \, \text{mol}Moles of C=12g/mol52.2g=4.35mol Moles of H=13.2 g1 g/mol=13.2 mol\text{Moles of H} = \frac{13.2 \, \text{g}}{1 \, \text{g/mol}} = 13.2 \, \text{mol}Moles of H=1g/mol13.2g=13.2mol Moles of O=34.6 g16 g/mol=2.16 mol\text{Moles of O} = \frac{34.6 \, \text{g}}{16 \, \text{g/mol}} = 2.16 \, \text{mol}Moles of O=16g/mol34.6g=2.16mol

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=4.35:13.2:2.16≈2:6:1\text{Ratio of C:H:O} = 4.35:13.2:2.16 \approx 2:6:1Ratio of C:H:O=4.35:13.2:2.16≈2:6:1

Step 4: Empirical formula:

Empirical formula=C2H6O\text{Empirical formula} = \text{C}_2\text{H}_6\text{O}Empirical formula=C2H6O

Explanation: Convert percentages to moles, then simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

94. How many moles of NaOH are needed to react with 2 moles of H₂SO₄?

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Use the stoichiometric ratio (1:2):

Moles of NaOH=2 mol H2SO4×2 mol NaOH1 mol H2SO4=4 mol\text{Moles of NaOH} = 2 \, \text{mol H}_2\text{SO}_4 \times \frac{2 \, \text{mol NaOH}}{1 \, \text{mol H}_2\text{SO}_4} = 4 \, \text{mol}Moles of NaOH=2mol H2SO4×1mol H2SO42mol NaOH=4mol

Explanation: Multiply the moles of the reactant by the stoichiometric coefficient to find the moles of the other reactant needed.

Concentration Terms (continued)

95. Calculate the volume of 0.3 M KCl solution needed to obtain 0.5 moles of KCl.

Volume=moles of solutemolarity=0.5 mol0.3 M=1.67 L\text{Volume} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.5 \, \text{mol}}{0.3 \, \text{M}} = 1.67 \, \text{L}Volume=molaritymoles of solute=0.3M0.5mol=1.67L

Explanation: Divide the number of moles of solute by the molarity to find the volume of the solution.

Relation Between Molarity and Normality (continued)

96. Find the normality of a 0.2 M NaOH solution.

NaOH→Na++OH−\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-NaOH→Na++OH−

Step 1: Determine the number of equivalents:

Number of OH− ions produced=1\text{Number of OH}^- \text{ ions produced} = 1Number of OH− ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.2 M×1=0.2 N\text{Normality} = 0.2 \, \text{M} \times 1 = 0.2 \, \text{N}Normality=0.2M×1=0.2N

Explanation: Normality is equal to molarity when the number of equivalents is 1.

Additional Numerical Problems with Explanations (continued)

97. Calculate the number of grams of KNO₃ required to prepare 1 liter of a 0.2 M solution.

Molecular weight of KNO3=39+14+3×16=101 g/mol\text{Molecular weight of KNO}_3 = 39 + 14 + 3 \times 16 = 101 \, \text{g/mol}Molecular weight of KNO3=39+14+3×16=101g/mol Moles of KNO3=0.2 M×1 L=0.2 mol\text{Moles of KNO}_3 = 0.2 \, \text{M} \times 1 \, \text{L} = 0.2 \, \text{mol}Moles of KNO3=0.2M×1L=0.2mol Mass of KNO3=0.2 mol×101 g/mol=20.2 g\text{Mass of KNO}_3 = 0.2 \, \text{mol} \times 101 \, \text{g/mol} = 20.2 \, \text{g}Mass of KNO3=0.2mol×101g/mol=20.2g

Explanation: Multiply the molarity by the volume to get the number of moles, then find the mass using the molecular weight.

98. How many liters of 0.5 M NaOH are needed to neutralize 250 mL of 0.2 M H₂SO₄?

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Determine moles of H₂SO₄:

Moles of H2SO4=0.2 M×0.25 L=0.05 mol\text{Moles of H}_2\text{SO}_4 = 0.2 \, \text{M} \times 0.25 \, \text{L} = 0.05 \, \text{mol}Moles of H2SO4=0.2M×0.25L=0.05mol

Step 2: Use the stoichiometric ratio (1:2) to find moles of NaOH:

Moles of NaOH=0.05 mol H2SO4×2 mol NaOH1 mol H2SO4=0.1 mol\text{Moles of NaOH} = 0.05 \, \text{mol H}_2\text{SO}_4 \times \frac{2 \, \text{mol NaOH}}{1 \, \text{mol H}_2\text{SO}_4} = 0.1 \, \text{mol}Moles of NaOH=0.05mol H2SO4×1mol H2SO42mol NaOH=0.1mol

Step 3: Calculate the volume of NaOH:

Volume of NaOH=moles of solutemolarity=0.1 mol0.5 M=0.2 L=200 mL\text{Volume of NaOH} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.1 \, \text{mol}}{0.5 \, \text{M}} = 0.2 \, \text{L} = 200 \, \text{mL}Volume of NaOH=molaritymoles of solute=0.5M0.1mol=0.2L=200mL

Explanation: Use the stoichiometric coefficients from the balanced equation and the molarity to determine the volume required for neutralization.

99. Determine the normality of a solution made by dissolving 25 g of HNO₃ in 500 mL of solution.

Molecular weight of HNO3=1+14+3×16=63 g/mol\text{Molecular weight of HNO}_3 = 1 + 14 + 3 \times 16 = 63 \, \text{g/mol}Molecular weight of HNO3=1+14+3×16=63g/mol Moles of HNO3=25 g63 g/mol=0.397 mol\text{Moles of HNO}_3 = \frac{25 \, \text{g}}{63 \, \text{g/mol}} = 0.397 \, \text{mol}Moles of HNO3=63g/mol25g=0.397mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=0.397 mol0.5 L=0.794 M\text{Molarity} = \frac{0.397 \, \text{mol}}{0.5 \, \text{L}} = 0.794 \, \text{M}Molarity=0.5L0.397mol=0.794M

Number of equivalents=1\text{Number of equivalents} = 1Number of equivalents=1 Normality=0.794 M×1=0.794 N\text{Normality} = 0.794 \, \text{M} \times 1 = 0.794 \, \text{N}Normality=0.794M×1=0.794N

Explanation: Calculate molarity first, then use the number of equivalents to find the normality.

100. Calculate the molality of a solution prepared by dissolving 50 g of Na₂SO₄ in 200 g of water.

Mass of solvent (water)=200 g=0.2 kg\text{Mass of solvent (water)} = 200 \, \text{g} = 0.2 \, \text{kg}Mass of solvent (water)=200g=0.2kg Molality=moles of solutemass of solvent in kg=0.352 mol0.2 kg=1.76 mol/kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.352 \, \text{mol}}{0.2 \, \text{kg}} = 1.76 \, \text{mol/kg}Molality=mass of solvent in kgmoles of solute=0.2kg0.352mol=1.76mol/kg

Explanation: Divide the number of moles of solute by the mass of solvent in kilograms to find the molality.

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Mole Concept (continued)

101. How many grams of O₂ are needed to react with 4 moles of C₄H₁₀ in a combustion reaction?

C4H10+O2→CO2+H2O\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}C4H10+O2→CO2+H2O

Balanced equation: 2C4H10+13O2→8CO2+10H2O2 \text{C}_4\text{H}_{10} + 13 \text{O}_2 \rightarrow 8 \text{CO}_2 + 10 \text{H}_2\text{O}2C4H10+13O2→8CO2+10H2O

Step 1: Moles of O₂ needed for 4 moles of C₄H₁₀:

Moles of O2=4 mol C4H10×13 mol O22 mol C4H10=26 mol O2\text{Moles of O}_2 = 4 \, \text{mol C}_4\text{H}_{10} \times \frac{13 \, \text{mol O}_2}{2 \, \text{mol C}_4\text{H}_{10}} = 26 \, \text{mol O}_2Moles of O2=4mol C4H10×2mol C4H1013mol O2=26mol O2

Step 2: Calculate the mass of O₂:

Molecular weight of O2=2×16=32 g/mol\text{Molecular weight of O}_2 = 2 \times 16 = 32 \, \text{g/mol}Molecular weight of O2=2×16=32g/mol Mass of O2=26 mol×32 g/mol=832 g\text{Mass of O}_2 = 26 \, \text{mol} \times 32 \, \text{g/mol} = 832 \, \text{g}Mass of O2=26mol×32g/mol=832g

Explanation: Use stoichiometric ratios to determine moles of O₂ required, then convert to mass.

102. Determine the number of molecules in 0.75 moles of H₂O.

Number of molecules=0.75 mol×6.022×1023 molecules/mol=4.5165×1023 molecules\text{Number of molecules} = 0.75 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 4.5165 \times 10^{23} \, \text{molecules}Number of molecules=0.75mol×6.022×1023molecules/mol=4.5165×1023molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

Determination of Formula of Compound (continued)

103. A compound contains 30% nitrogen and 70% oxygen by mass. Determine the empirical formula.

Step 1: Assume 100 g of the compound:

Nitrogen: 30 g
Oxygen: 70 g

Step 2: Convert mass to moles:

Moles of N=30 g14 g/mol=2.14 mol\text{Moles of N} = \frac{30 \, \text{g}}{14 \, \text{g/mol}} = 2.14 \, \text{mol}Moles of N=14g/mol30g=2.14mol Moles of O=70 g16 g/mol=4.38 mol\text{Moles of O} = \frac{70 \, \text{g}}{16 \, \text{g/mol}} = 4.38 \, \text{mol}Moles of O=16g/mol70g=4.38mol

Step 3: Determine the simplest mole ratio:

Ratio of N:O=2.14:4.38≈1:2\text{Ratio of N:O} = 2.14:4.38 \approx 1:2Ratio of N:O=2.14:4.38≈1:2

Step 4: Empirical formula:

Empirical formula=NO2\text{Empirical formula} = \text{NO}_2Empirical formula=NO2

Explanation: Convert percentages to moles, simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

104. How many grams of NaCl are required to react with 0.5 moles of AgNO₃?

NaCl+AgNO3→NaNO3+AgCl\text{NaCl} + \text{AgNO}_3 \rightarrow \text{NaNO}_3 + \text{AgCl}NaCl+AgNO3→NaNO3+AgCl

Step 1: Use stoichiometric ratios (1:1):

Moles of NaCl=0.5 mol AgNO3×1 mol NaCl1 mol AgNO3=0.5 mol\text{Moles of NaCl} = 0.5 \, \text{mol AgNO}_3 \times \frac{1 \, \text{mol NaCl}}{1 \, \text{mol AgNO}_3} = 0.5 \, \text{mol}Moles of NaCl=0.5mol AgNO3×1mol AgNO31mol NaCl=0.5mol

Step 2: Calculate the mass of NaCl:

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Mass of NaCl=0.5 mol×58.5 g/mol=29.25 g\text{Mass of NaCl} = 0.5 \, \text{mol} \times 58.5 \, \text{g/mol} = 29.25 \, \text{g}Mass of NaCl=0.5mol×58.5g/mol=29.25g

Explanation: Use stoichiometric ratios to determine moles of NaCl needed, then convert to mass.

Concentration Terms (continued)

105. Calculate the molarity of a solution prepared by dissolving 25 g of Na₂CO₃ in 750 mL of solution.

Molecular weight of Na2CO3=2×23+12+3×16=106 g/mol\text{Molecular weight of Na}_2\text{CO}_3 = 2 \times 23 + 12 + 3 \times 16 = 106 \, \text{g/mol}Molecular weight of Na2CO3=2×23+12+3×16=106g/mol Moles of Na2CO3=25 g106 g/mol=0.236 mol\text{Moles of Na}_2\text{CO}_3 = \frac{25 \, \text{g}}{106 \, \text{g/mol}} = 0.236 \, \text{mol}Moles of Na2CO3=106g/mol25g=0.236mol

Volume of solution=750 mL=0.75 L\text{Volume of solution} = 750 \, \text{mL} = 0.75 \, \text{L}Volume of solution=750mL=0.75L Molarity=0.236 mol0.75 L=0.315 M\text{Molarity} = \frac{0.236 \, \text{mol}}{0.75 \, \text{L}} = 0.315 \, \text{M}Molarity=0.75L0.236mol=0.315M

Explanation: Divide the number of moles of solute by the volume of solution in liters to find the molarity.

Relation Between Molarity and Normality (continued)

106. Determine the normality of a 0.5 M H₂SO₄ solution.

H2SO4→2H++SO42−\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}H2SO4→2H++SO42−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=2\text{Number of H}^+ \text{ ions produced} = 2Number of H+ ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×2=1.0 N\text{Normality} = 0.5 \, \text{M} \times 2 = 1.0 \, \text{N}Normality=0.5M×2=1.0N

Explanation: Multiply the molarity by the number of equivalents per mole of solute to get normality.

Additional Numerical Problems with Explanations (continued)

107. Calculate the volume of 1.5 M HCl needed to neutralize 200 mL of 0.2 M NaOH.

HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}HCl+NaOH→NaCl+H2O

Step 1: Determine moles of NaOH:

Moles of NaOH=0.2 M×0.2 L=0.04 mol\text{Moles of NaOH} = 0.2 \, \text{M} \times 0.2 \, \text{L} = 0.04 \, \text{mol}Moles of NaOH=0.2M×0.2L=0.04mol

Step 2: Use stoichiometric ratio (1:1) to find moles of HCl:

Moles of HCl=0.04 mol\text{Moles of HCl} = 0.04 \, \text{mol}Moles of HCl=0.04mol

Step 3: Calculate the volume of HCl:

Volume of HCl=moles of solutemolarity=0.04 mol1.5 M=0.027 L=27 mL\text{Volume of HCl} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.04 \, \text{mol}}{1.5 \, \text{M}} = 0.027 \, \text{L} = 27 \, \text{mL}Volume of HCl=molaritymoles of solute=1.5M0.04mol=0.027L=27mL

Explanation: Use stoichiometric ratios to determine moles of HCl needed, then calculate the volume using the molarity.

108. How many moles of KBr are present in 100 g of KBr?

Molecular weight of KBr=39+80=119 g/mol\text{Molecular weight of KBr} = 39 + 80 = 119 \, \text{g/mol}Molecular weight of KBr=39+80=119g/mol Moles of KBr=100 g119 g/mol=0.84 mol\text{Moles of KBr} = \frac{100 \, \text{g}}{119 \, \text{g/mol}} = 0.84 \, \text{mol}Moles of KBr=119g/mol100g=0.84mol

Explanation: Divide the mass of KBr by its molecular weight to determine the number of moles.

109. Find the molarity of a solution made by dissolving 15 g of NH₄Cl in 250 mL of solution.

Molecular weight of NH4Cl=14+4×1+35.5=53.5 g/mol\text{Molecular weight of NH}_4\text{Cl} = 14 + 4 \times 1 + 35.5 = 53.5 \, \text{g/mol}Molecular weight of NH4Cl=14+4×1+35.5=53.5g/mol Moles of NH4Cl=15 g53.5 g/mol=0.281 mol\text{Moles of NH}_4\text{Cl} = \frac{15 \, \text{g}}{53.5 \, \text{g/mol}} = 0.281 \, \text{mol}Moles of NH4Cl=53.5g/mol15g=0.281mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=0.281 mol0.25 L=1.12 M\text{Molarity} = \frac{0.281 \, \text{mol}}{0.25 \, \text{L}} = 1.12 \, \text{M}Molarity=0.25L0.281mol=1.12M

Explanation: Calculate the number of moles and divide by the volume in liters to find the molarity.

110. Calculate the normality of a solution prepared by dissolving 10 g of H₃PO₄ in 500 mL of solution.

Molecular weight of H3PO4=3×1+31+4×16=98 g/mol\text{Molecular weight of H}_3\text{PO}_4 = 3 \times 1 + 31 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H3PO4=3×1+31+4×16=98g/mol Moles of H3PO4=10 g98 g/mol=0.102 mol\text{Moles of H}_3\text{PO}_4 = \frac{10 \, \text{g}}{98 \, \text{g/mol}} = 0.102 \, \text{mol}Moles of H3PO4=98g/mol10g=0.102mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=0.102 mol0.5 L=0.204 M\text{Molarity} = \frac{0.102 \, \text{mol}}{0.5 \, \text{L}} = 0.204 \, \text{M}Molarity=0.5L0.102mol=0.204M

Number of equivalents=3\text{Number of equivalents} = 3Number of equivalents=3 Normality=0.204 M×3=0.612 N\text{Normality} = 0.204 \, \text{M} \times 3 = 0.612 \, \text{N}Normality=0.204M×3=0.612N

Explanation: Determine molarity, then multiply by the number of equivalents to find the normality.

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Mole Concept (continued)

111. How many moles of CO₂ are produced when 3 moles of C₆H₁₂O₆ undergo complete combustion?

Balanced equation:

C6H12O6+6O2→6CO2+6H2O\text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \rightarrow 6 \text{CO}_2 + 6 \text{H}_2\text{O}C6H12O6+6O2→6CO2+6H2O

Step 1: Use the stoichiometric ratio (1:6):

Moles of CO2=3 mol C6H12O6×6 mol CO21 mol C6H12O6=18 mol CO2\text{Moles of CO}_2 = 3 \, \text{mol C}_6\text{H}_{12}\text{O}_6 \times \frac{6 \, \text{mol CO}_2}{1 \, \text{mol C}_6\text{H}_{12}\text{O}_6} = 18 \, \text{mol CO}_2Moles of CO2=3mol C6H12O6×1mol C6H12O66mol CO2=18mol CO2

Explanation: Use the stoichiometric ratio to determine the number of moles of CO₂ produced.

112. Calculate the number of molecules in 0.25 moles of Na₂SO₄.

Number of molecules=0.25 mol×6.022×1023 molecules/mol=1.5055×1023 molecules\text{Number of molecules} = 0.25 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.5055 \times 10^{23} \, \text{molecules}Number of molecules=0.25mol×6.022×1023molecules/mol=1.5055×1023molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

Determination of Formula of Compound (continued)

113. A compound is found to contain 40% sulfur and 60% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Sulfur: 40 g
Oxygen: 60 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of S:O=1.25:3.75=1:3\text{Ratio of S:O} = 1.25:3.75 = 1:3Ratio of S:O=1.25:3.75=1:3

Step 4: Empirical formula:

Empirical formula=SO3\text{Empirical formula} = \text{SO}_3Empirical formula=SO3

Explanation: Convert percentages to moles, then simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

114. How many grams of CO₂ are produced when 2 moles of C₃H₈ are burned in excess oxygen?

Balanced equation:

C3H8+5O2→3CO2+4H2O\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}C3H8+5O2→3CO2+4H2O

Step 1: Use the stoichiometric ratio (1:3):

Moles of CO2=2 mol C3H8×3 mol CO21 mol C3H8=6 mol CO2\text{Moles of CO}_2 = 2 \, \text{mol C}_3\text{H}_8 \times \frac{3 \, \text{mol CO}_2}{1 \, \text{mol C}_3\text{H}_8} = 6 \, \text{mol CO}_2Moles of CO2=2mol C3H8×1mol C3H83mol CO2=6mol CO2

Step 2: Calculate the mass of CO₂:

Molecular weight of CO2=12+2×16=44 g/mol\text{Molecular weight of CO}_2 = 12 + 2 \times 16 = 44 \, \text{g/mol}Molecular weight of CO2=12+2×16=44g/mol Mass of CO2=6 mol×44 g/mol=264 g\text{Mass of CO}_2 = 6 \, \text{mol} \times 44 \, \text{g/mol} = 264 \, \text{g}Mass of CO2=6mol×44g/mol=264g

Explanation: Use stoichiometric ratios to determine moles of CO₂ produced, then convert to mass.

Concentration Terms (continued)

115. Calculate the volume of a 0.4 M NaCl solution required to prepare 1 liter of a 0.1 M NaCl solution.

Step 1: Use the dilution formula:

C1V1=C2V2C_1V_1 = C_2V_2C1V1=C2V2

where C1C_1C1 is the initial concentration, V1V_1V1 is the initial volume, C2C_2C2 is the final concentration, and V2V_2V2 is the final volume.

0.4 M×V1=0.1 M×1 L0.4 \, \text{M} \times V_1 = 0.1 \, \text{M} \times 1 \, \text{L}0.4M×V1=0.1M×1L V1=0.1 M×1 L0.4 M=0.25 L=250 mLV_1 = \frac{0.1 \, \text{M} \times 1 \, \text{L}}{0.4 \, \text{M}} = 0.25 \, \text{L} = 250 \, \text{mL}V1=0.4M0.1M×1L=0.25L=250mL

Explanation: Use the dilution formula to find the volume of the more concentrated solution needed.

Relation Between Molarity and Normality (continued)

116. Find the normality of a 0.6 M H₃PO₄ solution.

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3\text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=3\text{Number of H}^+ \text{ ions produced} = 3Number of H+ ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.6 M×3=1.8 N\text{Normality} = 0.6 \, \text{M} \times 3 = 1.8 \, \text{N}Normality=0.6M×3=1.8N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

117. How many moles of Na₂CO₃ are in 200 g of Na₂CO₃?

Molecular weight of Na2CO3=2×23+12+3×16=106 g/mol\text{Molecular weight of Na}_2\text{CO}_3 = 2 \times 23 + 12 + 3 \times 16 = 106 \, \text{g/mol}Molecular weight of Na2CO3=2×23+12+3×16=106g/mol Moles of Na2CO3=200 g106 g/mol=1.89 mol\text{Moles of Na}_2\text{CO}_3 = \frac{200 \, \text{g}}{106 \, \text{g/mol}} = 1.89 \, \text{mol}Moles of Na2CO3=106g/mol200g=1.89mol

Explanation: Divide the mass of Na₂CO₃ by its molecular weight to determine the number of moles.

118. Calculate the volume of 0.8 M HCl required to react with 50 mL of 0.2 M NaOH.

HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}HCl+NaOH→NaCl+H2O

Step 1: Determine moles of NaOH:

Moles of NaOH=0.2 M×0.05 L=0.01 mol\text{Moles of NaOH} = 0.2 \, \text{M} \times 0.05 \, \text{L} = 0.01 \, \text{mol}Moles of NaOH=0.2M×0.05L=0.01mol

Step 2: Use stoichiometric ratio (1:1) to find moles of HCl:

Moles of HCl=0.01 mol\text{Moles of HCl} = 0.01 \, \text{mol}Moles of HCl=0.01mol

Step 3: Calculate the volume of HCl:

Volume of HCl=moles of solutemolarity=0.01 mol0.8 M=0.0125 L=12.5 mL\text{Volume of HCl} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.01 \, \text{mol}}{0.8 \, \text{M}} = 0.0125 \, \text{L} = 12.5 \, \text{mL}Volume of HCl=molaritymoles of solute=0.8M0.01mol=0.0125L=12.5mL

Explanation: Determine the moles of NaOH and use stoichiometry to find moles of HCl needed, then convert to volume.

119. Determine the number of moles in 75 g of CCl₄.

Molecular weight of CCl4=12+4×35.5=154 g/mol\text{Molecular weight of CCl}_4 = 12 + 4 \times 35.5 = 154 \, \text{g/mol}Molecular weight of CCl4=12+4×35.5=154g/mol Moles of CCl4=75 g154 g/mol=0.487 mol\text{Moles of CCl}_4 = \frac{75 \, \text{g}}{154 \, \text{g/mol}} = 0.487 \, \text{mol}Moles of CCl4=154g/mol75g=0.487mol

Explanation: Divide the mass of CCl₄ by its molecular weight to find the number of moles.

120. Calculate the molarity of a solution prepared by dissolving 10 g of K₂SO₄ in 200 mL of solution.

Molecular weight of K2SO4=2×39+32+4×16=174 g/mol\text{Molecular weight of K}_2\text{SO}_4 = 2 \times 39 + 32 + 4 \times 16 = 174 \, \text{g/mol}Molecular weight of K2SO4=2×39+32+4×16=174g/mol Moles of K2SO4=10 g174 g/mol=0.0575 mol\text{Moles of K}_2\text{SO}_4 = \frac{10 \, \text{g}}{174 \, \text{g/mol}} = 0.0575 \, \text{mol}Moles of K2SO4=174g/mol10g=0.0575mol

Volume of solution=200 mL=0.2 L\text{Volume of solution} = 200 \, \text{mL} = 0.2 \, \text{L}Volume of solution=200mL=0.2L Molarity=0.0575 mol0.2 L=0.2875 M\text{Molarity} = \frac{0.0575 \, \text{mol}}{0.2 \, \text{L}} = 0.2875 \, \text{M}Molarity=0.2L0.0575mol=0.2875M

Explanation: Calculate the number of moles and divide by the volume in liters to find the molarity.

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Mole Concept (continued)

121. How many grams of H₂SO₄ are required to produce 0.5 moles of H₂O?

Balanced equation:

H2SO4→H2O+SO3\text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{SO}_3H2SO4→H2O+SO3

Step 1: Determine the number of moles of H₂SO₄ needed (1 mole of H₂SO₄ produces 1 mole of H₂O):

Moles of H2SO4=0.5 mol\text{Moles of H}_2\text{SO}_4 = 0.5 \, \text{mol}Moles of H2SO4=0.5mol

Step 2: Calculate the mass of H₂SO₄:

Molecular weight of H2SO4=2×1+32+4×16=98 g/mol\text{Molecular weight of H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}Molecular weight of H2SO4=2×1+32+4×16=98g/mol Mass of H2SO4=0.5 mol×98 g/mol=49 g\text{Mass of H}_2\text{SO}_4 = 0.5 \, \text{mol} \times 98 \, \text{g/mol} = 49 \, \text{g}Mass of H2SO4=0.5mol×98g/mol=49g

Explanation: Use stoichiometric ratios to find the amount of H₂SO₄ needed, then convert to mass.

122. Calculate the number of molecules in 3 moles of NaCl.

Number of molecules=3 mol×6.022×1023 molecules/mol=1.8066×1024 molecules\text{Number of molecules} = 3 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.8066 \times 10^{24} \, \text{molecules}Number of molecules=3mol×6.022×1023molecules/mol=1.8066×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

Determination of Formula of Compound (continued)

123. A compound contains 20% nitrogen and 80% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Nitrogen: 20 g
Oxygen: 80 g

Step 2: Convert mass to moles:

Moles of N=20 g14 g/mol=1.43 mol\text{Moles of N} = \frac{20 \, \text{g}}{14 \, \text{g/mol}} = 1.43 \, \text{mol}Moles of N=14g/mol20g=1.43mol Moles of O=80 g16 g/mol=5.0 mol\text{Moles of O} = \frac{80 \, \text{g}}{16 \, \text{g/mol}} = 5.0 \, \text{mol}Moles of O=16g/mol80g=5.0mol

Step 3: Determine the simplest mole ratio:

Ratio of N:O=1.43:5.0=1:3.5≈2:7\text{Ratio of N:O} = 1.43:5.0 = 1:3.5 \approx 2:7Ratio of N:O=1.43:5.0=1:3.5≈2:7

Step 4: Empirical formula:

Empirical formula=N2O7\text{Empirical formula} = \text{N}_2\text{O}_7Empirical formula=N2O7

Explanation: Convert percentages to moles, simplify the mole ratio to determine the empirical formula.

Stoichiometric Calculations (continued)

124. How many grams of H₂O are produced when 4 moles of Al react with excess HCl?

Balanced equation:

2Al+6HCl→2AlCl3+3H2O2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2\text{O}2Al+6HCl→2AlCl3+3H2O

Step 1: Use stoichiometric ratio (2:3):

Moles of H2O=4 mol Al×3 mol H2O2 mol Al=6 mol H2O\text{Moles of H}_2\text{O} = 4 \, \text{mol Al} \times \frac{3 \, \text{mol H}_2\text{O}}{2 \, \text{mol Al}} = 6 \, \text{mol H}_2\text{O}Moles of H2O=4mol Al×2mol Al3mol H2O=6mol H2O

Step 2: Calculate the mass of H₂O:

Molecular weight of H2O=2×1+16=18 g/mol\text{Molecular weight of H}_2\text{O} = 2 \times 1 + 16 = 18 \, \text{g/mol}Molecular weight of H2O=2×1+16=18g/mol Mass of H2O=6 mol×18 g/mol=108 g\text{Mass of H}_2\text{O} = 6 \, \text{mol} \times 18 \, \text{g/mol} = 108 \, \text{g}Mass of H2O=6mol×18g/mol=108g

Explanation: Use stoichiometric ratios to determine moles of H₂O produced, then convert to mass.

Concentration Terms (continued)

125. Calculate the molarity of a solution prepared by dissolving 12 g of KNO₃ in 300 mL of solution.

Volume of solution=300 mL=0.3 L\text{Volume of solution} = 300 \, \text{mL} = 0.3 \, \text{L}Volume of solution=300mL=0.3L Molarity=0.119 mol0.3 L=0.397 M\text{Molarity} = \frac{0.119 \, \text{mol}}{0.3 \, \text{L}} = 0.397 \, \text{M}Molarity=0.3L0.119mol=0.397M

Explanation: Calculate the number of moles and divide by the volume in liters to find the molarity.

Relation Between Molarity and Normality (continued)

126. Determine the normality of a 0.5 M HNO₃ solution.

HNO3→H++NO3−\text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^-HNO3→H++NO3−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=1\text{Number of H}^+ \text{ ions produced} = 1Number of H+ ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×1=0.5 N\text{Normality} = 0.5 \, \text{M} \times 1 = 0.5 \, \text{N}Normality=0.5M×1=0.5N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

127. Calculate the number of moles in 50 g of C₁₂H₂₂O₁₁ (sucrose).

Molecular weight of C12H22O11=12×12+22×1+11×16=342 g/mol\text{Molecular weight of C}_{12}\text{H}_{22}\text{O}_{11} = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \, \text{g/mol}Molecular weight of C12H22O11=12×12+22×1+11×16=342g/mol Moles of sucrose=50 g342 g/mol=0.146 mol\text{Moles of sucrose} = \frac{50 \, \text{g}}{342 \, \text{g/mol}} = 0.146 \, \text{mol}Moles of sucrose=342g/mol50g=0.146mol

Explanation: Divide the mass of sucrose by its molecular weight to determine the number of moles.

128. Calculate the volume of a 0.5 M NaOH solution required to neutralize 75 mL of 0.2 M H₂SO₄.

Balanced equation:

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Determine moles of H₂SO₄:

Moles of H2SO4=0.2 M×0.075 L=0.015 mol\text{Moles of H}_2\text{SO}_4 = 0.2 \, \text{M} \times 0.075 \, \text{L} = 0.015 \, \text{mol}Moles of H2SO4=0.2M×0.075L=0.015mol

Step 2: Use stoichiometric ratio (1:2) to find moles of NaOH:

Moles of NaOH=0.015 mol H2SO4×2 mol NaOH1 mol H2SO4=0.03 mol NaOH\text{Moles of NaOH} = 0.015 \, \text{mol H}_2\text{SO}_4 \times \frac{2 \, \text{mol NaOH}}{1 \, \text{mol H}_2\text{SO}_4} = 0.03 \, \text{mol NaOH}Moles of NaOH=0.015mol H2SO4×1mol H2SO42mol NaOH=0.03mol NaOH

Step 3: Calculate the volume of NaOH:

Volume of NaOH=moles of solutemolarity=0.03 mol0.5 M=0.06 L=60 mL\text{Volume of NaOH} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.03 \, \text{mol}}{0.5 \, \text{M}} = 0.06 \, \text{L} = 60 \, \text{mL}Volume of NaOH=molaritymoles of solute=0.5M0.03mol=0.06L=60mL

Explanation: Use stoichiometric ratios to determine moles of NaOH needed, then convert to volume using the molarity.

129. Determine the molarity of a solution prepared by dissolving 8 g of Na₂SO₄ in 500 mL of solution.

Molecular weight of Na2SO4=2×23+32+4×16=174 g/mol\text{Molecular weight of Na}_2\text{SO}_4 = 2 \times 23 + 32 + 4 \times 16 = 174 \, \text{g/mol}Molecular weight of Na2SO4=2×23+32+4×16=174g/mol Moles of Na2SO4=8 g174 g/mol=0.046 mol\text{Moles of Na}_2\text{SO}_4 = \frac{8 \, \text{g}}{174 \, \text{g/mol}} = 0.046 \, \text{mol}Moles of Na2SO4=174g/mol8g=0.046mol

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Mole Concept (continued)

129. Determine the molarity of a solution prepared by dissolving 8 g of Na₂SO₄ in 500 mL of solution.

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=0.046 mol0.5 L=0.092 M\text{Molarity} = \frac{0.046 \, \text{mol}}{0.5 \, \text{L}} = 0.092 \, \text{M}Molarity=0.5L0.046mol=0.092M

Explanation: Calculate the number of moles and divide by the volume in liters to find the molarity.

130. How many moles of KCl are there in 250 mL of a 0.4 M KCl solution?

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Moles of KCl=0.4 M×0.25 L=0.1 mol\text{Moles of KCl} = 0.4 \, \text{M} \times 0.25 \, \text{L} = 0.1 \, \text{mol}Moles of KCl=0.4M×0.25L=0.1mol

Explanation: Multiply the molarity by the volume in liters to determine the number of moles.

Determination of Formula of Compound (continued)

131. A compound is found to contain 30% nitrogen and 70% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Nitrogen: 30 g
Oxygen: 70 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of N:O=2.14:4.38≈1:2\text{Ratio of N:O} = 2.14:4.38 \approx 1:2Ratio of N:O=2.14:4.38≈1:2

Step 4: Empirical formula:

Empirical formula=NO2\text{Empirical formula} = \text{NO}_2Empirical formula=NO2

Explanation: Convert percentages to moles, then simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

132. How many grams of CO are produced when 5 moles of Fe react with excess CO₂?

Balanced equation:

2Fe+3CO2→2FeO+3CO2 \text{Fe} + 3 \text{CO}_2 \rightarrow 2 \text{FeO} + 3 \text{CO}2Fe+3CO2→2FeO+3CO

Step 1: Use stoichiometric ratio (2:3):

Moles of CO=5 mol Fe×3 mol CO2 mol Fe=7.5 mol CO\text{Moles of CO} = 5 \, \text{mol Fe} \times \frac{3 \, \text{mol CO}}{2 \, \text{mol Fe}} = 7.5 \, \text{mol CO}Moles of CO=5mol Fe×2mol Fe3mol CO=7.5mol CO

Step 2: Calculate the mass of CO:

Molecular weight of CO=12+16=28 g/mol\text{Molecular weight of CO} = 12 + 16 = 28 \, \text{g/mol}Molecular weight of CO=12+16=28g/mol Mass of CO=7.5 mol×28 g/mol=210 g\text{Mass of CO} = 7.5 \, \text{mol} \times 28 \, \text{g/mol} = 210 \, \text{g}Mass of CO=7.5mol×28g/mol=210g

Explanation: Use stoichiometric ratios to determine the moles of CO produced, then convert to mass.

Concentration Terms (continued)

133. Calculate the volume of 0.6 M H₂SO₄ required to prepare 2 liters of 0.2 M H₂SO₄ solution.

Step 1: Use the dilution formula:

C1V1=C2V2C_1V_1 = C_2V_2C1V1=C2V2

where C1C_1C1 is the initial concentration, V1V_1V1 is the initial volume, C2C_2C2 is the final concentration, and V2V_2V2 is the final volume.

0.6 M×V1=0.2 M×2 L0.6 \, \text{M} \times V_1 = 0.2 \, \text{M} \times 2 \, \text{L}0.6M×V1=0.2M×2L V1=0.2 M×2 L0.6 M=0.4 L0.6=0.667 L=667 mLV_1 = \frac{0.2 \, \text{M} \times 2 \, \text{L}}{0.6 \, \text{M}} = \frac{0.4 \, \text{L}}{0.6} = 0.667 \, \text{L} = 667 \, \text{mL}V1=0.6M0.2M×2L=0.60.4L=0.667L=667mL

Explanation: Use the dilution formula to determine the volume of the more concentrated solution needed.

Relation Between Molarity and Normality (continued)

134. Find the normality of a 0.3 M NaOH solution.

NaOH→Na++OH−\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-NaOH→Na++OH−

Step 1: Determine the number of equivalents:

Number of OH− ions produced=1\text{Number of OH}^- \text{ ions produced} = 1Number of OH− ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.3 M×1=0.3 N\text{Normality} = 0.3 \, \text{M} \times 1 = 0.3 \, \text{N}Normality=0.3M×1=0.3N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

135. Calculate the number of moles in 50 g of KBr.

Molecular weight of KBr=39+80=119 g/mol\text{Molecular weight of KBr} = 39 + 80 = 119 \, \text{g/mol}Molecular weight of KBr=39+80=119g/mol Moles of KBr=50 g119 g/mol=0.42 mol\text{Moles of KBr} = \frac{50 \, \text{g}}{119 \, \text{g/mol}} = 0.42 \, \text{mol}Moles of KBr=119g/mol50g=0.42mol

Explanation: Divide the mass of KBr by its molecular weight to determine the number of moles.

136. Calculate the volume of 0.2 M Na₂SO₄ solution needed to provide 0.5 moles of Na₂SO₄.

Volume of solution=moles of solutemolarity=0.5 mol0.2 M=2.5 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.5 \, \text{mol}}{0.2 \, \text{M}} = 2.5 \, \text{L}Volume of solution=molaritymoles of solute=0.2M0.5mol=2.5L

Explanation: Use the molarity formula to calculate the volume of solution required.

137. How many grams of NaOH are needed to prepare 500 mL of a 0.4 M NaOH solution?

Moles of NaOH=0.4 M×0.5 L=0.2 mol\text{Moles of NaOH} = 0.4 \, \text{M} \times 0.5 \, \text{L} = 0.2 \, \text{mol}Moles of NaOH=0.4M×0.5L=0.2mol Molecular weight of NaOH=23+16+1=40 g/mol\text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}Molecular weight of NaOH=23+16+1=40g/mol Mass of NaOH=0.2 mol×40 g/mol=8 g\text{Mass of NaOH} = 0.2 \, \text{mol} \times 40 \, \text{g/mol} = 8 \, \text{g}Mass of NaOH=0.2mol×40g/mol=8g

Explanation: Calculate the moles of NaOH required and convert to mass.

138. Calculate the number of molecules in 2 moles of C₂H₄.

Number of molecules=2 mol×6.022×1023 molecules/mol=1.2044×1024 molecules\text{Number of molecules} = 2 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.2044 \times 10^{24} \, \text{molecules}Number of molecules=2mol×6.022×1023molecules/mol=1.2044×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

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Mole Concept (continued)

139. How many grams of NH₄Cl are required to make 250 mL of a 0.5 M solution?

Step 1: Calculate the moles of NH₄Cl required:

Moles of NH4Cl=0.5 M×0.25 L=0.125 mol\text{Moles of NH}_4\text{Cl} = 0.5 \, \text{M} \times 0.25 \, \text{L} = 0.125 \, \text{mol}Moles of NH4Cl=0.5M×0.25L=0.125mol

Step 2: Calculate the mass of NH₄Cl:

Molecular weight of NH4Cl=14+4×1+35.5=53.5 g/mol\text{Molecular weight of NH}_4\text{Cl} = 14 + 4 \times 1 + 35.5 = 53.5 \, \text{g/mol}Molecular weight of NH4Cl=14+4×1+35.5=53.5g/mol Mass of NH4Cl=0.125 mol×53.5 g/mol=6.69 g\text{Mass of NH}_4\text{Cl} = 0.125 \, \text{mol} \times 53.5 \, \text{g/mol} = 6.69 \, \text{g}Mass of NH4Cl=0.125mol×53.5g/mol=6.69g

Explanation: Calculate the number of moles needed and convert it to mass using the molecular weight.

140. Determine the molarity of a solution prepared by dissolving 15 g of NaHCO₃ in 250 mL of solution.

Molecular weight of NaHCO3=23+1+12+3×16=84 g/mol\text{Molecular weight of NaHCO}_3 = 23 + 1 + 12 + 3 \times 16 = 84 \, \text{g/mol}Molecular weight of NaHCO3=23+1+12+3×16=84g/mol Moles of NaHCO3=15 g84 g/mol=0.179 mol\text{Moles of NaHCO}_3 = \frac{15 \, \text{g}}{84 \, \text{g/mol}} = 0.179 \, \text{mol}Moles of NaHCO3=84g/mol15g=0.179mol

Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=0.179 mol0.25 L=0.716 M\text{Molarity} = \frac{0.179 \, \text{mol}}{0.25 \, \text{L}} = 0.716 \, \text{M}Molarity=0.25L0.179mol=0.716M

Explanation: Convert the mass of NaHCO₃ to moles and then find the molarity by dividing by the volume in liters.

Determination of Formula of Compound (continued)

141. A compound is composed of 40% phosphorus and 60% oxygen by mass. Find its empirical formula.

Step 1: Assume 100 g of the compound:

Phosphorus: 40 g
Oxygen: 60 g

Step 2: Convert mass to moles:

Moles of P=40 g31 g/mol=1.29 mol\text{Moles of P} = \frac{40 \, \text{g}}{31 \, \text{g/mol}} = 1.29 \, \text{mol}Moles of P=31g/mol40g=1.29mol Moles of O=60 g16 g/mol=3.75 mol\text{Moles of O} = \frac{60 \, \text{g}}{16 \, \text{g/mol}} = 3.75 \, \text{mol}Moles of O=16g/mol60g=3.75mol

Step 3: Determine the simplest mole ratio:

Ratio of P:O=1.29:3.75≈1:2.9≈1:3\text{Ratio of P:O} = 1.29:3.75 \approx 1:2.9 \approx 1:3Ratio of P:O=1.29:3.75≈1:2.9≈1:3

Step 4: Empirical formula:

Empirical formula=PO4\text{Empirical formula} = \text{PO}_4Empirical formula=PO4

Explanation: Convert percentages to moles, simplify the mole ratio to get the empirical formula.

Stoichiometric Calculations (continued)

142. How many moles of NaCl are needed to completely react with 0.5 moles of AgNO₃?

Balanced equation:

NaCl+AgNO3→NaNO3+AgCl\text{NaCl} + \text{AgNO}_3 \rightarrow \text{NaNO}_3 + \text{AgCl}NaCl+AgNO3→NaNO3+AgCl

Step 1: Use stoichiometric ratio (1:1):

Moles of NaCl=0.5 mol AgNO3×1 mol NaCl1 mol AgNO3=0.5 mol NaCl\text{Moles of NaCl} = 0.5 \, \text{mol AgNO}_3 \times \frac{1 \, \text{mol NaCl}}{1 \, \text{mol AgNO}_3} = 0.5 \, \text{mol NaCl}Moles of NaCl=0.5mol AgNO3×1mol AgNO31mol NaCl=0.5mol NaCl

Explanation: Use the stoichiometric ratio to find the moles of NaCl required for the reaction.

Concentration Terms (continued)

143. Calculate the normality of a 0.8 M H₃PO₄ solution.

Balanced equation:

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3 \text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=3\text{Number of H}^+ \text{ ions produced} = 3Number of H+ ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.8 M×3=2.4 N\text{Normality} = 0.8 \, \text{M} \times 3 = 2.4 \, \text{N}Normality=0.8M×3=2.4N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Relation Between Molarity and Normality (continued)

144. Find the molarity of a solution prepared by diluting 100 mL of 2 N HCl to a final volume of 1 liter.

Step 1: Use the dilution formula:

C1V1=C2V2C_1V_1 = C_2V_2C1V1=C2V2

where C1C_1C1 is the initial normality, V1V_1V1 is the initial volume, C2C_2C2 is the final molarity, and V2V_2V2 is the final volume.

2 N×0.1 L=C2×1 L2 \, \text{N} \times 0.1 \, \text{L} = C_2 \times 1 \, \text{L}2N×0.1L=C2×1L C2=2×0.11=0.2 MC_2 = \frac{2 \times 0.1}{1} = 0.2 \, \text{M}C2=12×0.1=0.2M

Explanation: Use the dilution formula to find the new molarity after dilution.

Additional Numerical Problems with Explanations (continued)

145. Calculate the volume of 0.5 M K₂SO₄ solution required to provide 0.2 moles of K₂SO₄.

Volume of solution=moles of solutemolarity=0.2 mol0.5 M=0.4 L=400 mL\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.2 \, \text{mol}}{0.5 \, \text{M}} = 0.4 \, \text{L} = 400 \, \text{mL}Volume of solution=molaritymoles of solute=0.5M0.2mol=0.4L=400mL

Explanation: Use the molarity formula to calculate the volume of solution needed.

146. How many grams of Na₂CO₃ are required to make 1 liter of a 0.25 M solution?

Moles of Na2CO3=0.25 M×1 L=0.25 mol\text{Moles of Na}_2\text{CO}_3 = 0.25 \, \text{M} \times 1 \, \text{L} = 0.25 \, \text{mol}Moles of Na2CO3=0.25M×1L=0.25mol Molecular weight of Na2CO3=2×23+12+3×16=106 g/mol\text{Molecular weight of Na}_2\text{CO}_3 = 2 \times 23 + 12 + 3 \times 16 = 106 \, \text{g/mol}Molecular weight of Na2CO3=2×23+12+3×16=106g/mol Mass of Na2CO3=0.25 mol×106 g/mol=26.5 g\text{Mass of Na}_2\text{CO}_3 = 0.25 \, \text{mol} \times 106 \, \text{g/mol} = 26.5 \, \text{g}Mass of Na2CO3=0.25mol×106g/mol=26.5g

Explanation: Calculate the moles needed and convert to mass using the molecular weight.

147. Calculate the number of moles in 100 mL of a 0.8 M Na₂SO₄ solution.

Volume of solution=100 mL=0.1 L\text{Volume of solution} = 100 \, \text{mL} = 0.1 \, \text{L}Volume of solution=100mL=0.1L Moles of Na2SO4=0.8 M×0.1 L=0.08 mol\text{Moles of Na}_2\text{SO}_4 = 0.8 \, \text{M} \times 0.1 \, \text{L} = 0.08 \, \text{mol}Moles of Na2SO4=0.8M×0.1L=0.08mol

Explanation: Multiply the molarity by the volume in liters to determine the number of moles.

148. Determine the number of molecules in 4 moles of CO₂.

Number of molecules=4 mol×6.022×1023 molecules/mol=2.4088×1024 molecules\text{Number of molecules} = 4 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 2.4088 \times 10^{24} \, \text{molecules}Number of molecules=4mol×6.022×1023molecules/mol=2.4088×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

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Mole Concept (continued)

149. Calculate the number of moles of NaCl in 75 g of the compound.

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Moles of NaCl=75 g58.5 g/mol=1.28 mol\text{Moles of NaCl} = \frac{75 \, \text{g}}{58.5 \, \text{g/mol}} = 1.28 \, \text{mol}Moles of NaCl=58.5g/mol75g=1.28mol

Explanation: Divide the mass of NaCl by its molecular weight to determine the number of moles.

150. What is the molarity of a solution if 10 g of Na₂SO₄ is dissolved in 200 mL of solution?

Molecular weight of Na2SO4=2×23+32+4×16=174 g/mol\text{Molecular weight of Na}_2\text{SO}_4 = 2 \times 23 + 32 + 4 \times 16 = 174 \, \text{g/mol}Molecular weight of Na2SO4=2×23+32+4×16=174g/mol Moles of Na2SO4=10 g174 g/mol=0.0575 mol\text{Moles of Na}_2\text{SO}_4 = \frac{10 \, \text{g}}{174 \, \text{g/mol}} = 0.0575 \, \text{mol}Moles of Na2SO4=174g/mol10g=0.0575mol

Volume of solution=200 mL=0.2 L\text{Volume of solution} = 200 \, \text{mL} = 0.2 \, \text{L}Volume of solution=200mL=0.2L Molarity=0.0575 mol0.2 L=0.288 M\text{Molarity} = \frac{0.0575 \, \text{mol}}{0.2 \, \text{L}} = 0.288 \, \text{M}Molarity=0.2L0.0575mol=0.288M

Explanation: Convert mass to moles and then divide by the volume in liters to find the molarity.

Determination of Formula of Compound (continued)

151. A compound contains 20% carbon, 6.67% hydrogen, and 73.33% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 20 g
Hydrogen: 6.67 g
Oxygen: 73.33 g

Step 2: Convert mass to moles:

Moles of C=20 g12 g/mol=1.67 mol\text{Moles of C} = \frac{20 \, \text{g}}{12 \, \text{g/mol}} = 1.67 \, \text{mol}Moles of C=12g/mol20g=1.67mol Moles of H=6.67 g1 g/mol=6.67 mol\text{Moles of H} = \frac{6.67 \, \text{g}}{1 \, \text{g/mol}} = 6.67 \, \text{mol}Moles of H=1g/mol6.67g=6.67mol Moles of O=73.33 g16 g/mol=4.58 mol\text{Moles of O} = \frac{73.33 \, \text{g}}{16 \, \text{g/mol}} = 4.58 \, \text{mol}Moles of O=16g/mol73.33g=4.58mol

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=1.67:6.67:4.58≈1:4:3\text{Ratio of C:H:O} = 1.67:6.67:4.58 \approx 1:4:3Ratio of C:H:O=1.67:6.67:4.58≈1:4:3

Step 4: Empirical formula:

Empirical formula=C1H4O3\text{Empirical formula} = \text{C}_1\text{H}_4\text{O}_3Empirical formula=C1H4O3

Explanation: Convert percentages to moles, simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

152. How many liters of 0.5 M H₂SO₄ are needed to react completely with 1 mole of Na₂CO₃?

Balanced equation:

H2SO4+Na2CO3→Na2SO4+H2O+CO2\text{H}_2\text{SO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2H2SO4+Na2CO3→Na2SO4+H2O+CO2

Step 1: Use stoichiometric ratio (1:1):

Moles of H2SO4=1 mol Na2CO3×1 mol H2SO41 mol Na2CO3=1 mol H2SO4\text{Moles of H}_2\text{SO}_4 = 1 \, \text{mol Na}_2\text{CO}_3 \times \frac{1 \, \text{mol H}_2\text{SO}_4}{1 \, \text{mol Na}_2\text{CO}_3} = 1 \, \text{mol H}_2\text{SO}_4Moles of H2SO4=1mol Na2CO3×1mol Na2CO31mol H2SO4=1mol H2SO4

Step 2: Calculate the volume of 0.5 M H₂SO₄ solution:

Volume of solution=moles of solutemolarity=1 mol0.5 M=2 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{1 \, \text{mol}}{0.5 \, \text{M}} = 2 \, \text{L}Volume of solution=molaritymoles of solute=0.5M1mol=2L

Explanation: Use the stoichiometric ratio and molarity to determine the volume of solution required.

Concentration Terms (continued)

153. What is the normality of a solution if 25 mL of 0.2 M H₂SO₄ is diluted to 250 mL?

Step 1: Use the dilution formula:

C1V1=C2V2C_1V_1 = C_2V_2C1V1=C2V2

where C1C_1C1 is the initial concentration, V1V_1V1 is the initial volume, C2C_2C2 is the final concentration, and V2V_2V2 is the final volume.

Step 2: Calculate the final molarity:

0.2 M×0.025 L=C2×0.25 L0.2 \, \text{M} \times 0.025 \, \text{L} = C_2 \times 0.25 \, \text{L}0.2M×0.025L=C2×0.25L C2=0.2×0.0250.25=0.02 MC_2 = \frac{0.2 \times 0.025}{0.25} = 0.02 \, \text{M}C2=0.250.2×0.025=0.02M

Step 3: Convert molarity to normality:

Number of H+ ions produced=2\text{Number of H}^+ \text{ ions produced} = 2Number of H+ ions produced=2 Normality=0.02 M×2=0.04 N\text{Normality} = 0.02 \, \text{M} \times 2 = 0.04 \, \text{N}Normality=0.02M×2=0.04N

Explanation: First, find the final molarity after dilution and then convert it to normality.

Relation Between Molarity and Normality (continued)

154. Calculate the normality of a 0.5 M HNO₃ solution.

Balanced equation:

HNO3→H++NO3−\text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^-HNO3→H++NO3−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=1\text{Number of H}^+ \text{ ions produced} = 1Number of H+ ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×1=0.5 N\text{Normality} = 0.5 \, \text{M} \times 1 = 0.5 \, \text{N}Normality=0.5M×1=0.5N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

155. How many grams of MgCl₂ are required to make 500 mL of a 0.2 M solution?

Moles of MgCl2=0.2 M×0.5 L=0.1 mol\text{Moles of MgCl}_2 = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{mol}Moles of MgCl2=0.2M×0.5L=0.1mol Molecular weight of MgCl2=24+2×35.5=95 g/mol\text{Molecular weight of MgCl}_2 = 24 + 2 \times 35.5 = 95 \, \text{g/mol}Molecular weight of MgCl2=24+2×35.5=95g/mol Mass of MgCl2=0.1 mol×95 g/mol=9.5 g\text{Mass of MgCl}_2 = 0.1 \, \text{mol} \times 95 \, \text{g/mol} = 9.5 \, \text{g}Mass of MgCl2=0.1mol×95g/mol=9.5g

Explanation: Calculate the moles required and then convert it to mass using the molecular weight.

156. Determine the number of moles in 400 mL of a 0.6 M KNO₃ solution.

Volume of solution=400 mL=0.4 L\text{Volume of solution} = 400 \, \text{mL} = 0.4 \, \text{L}Volume of solution=400mL=0.4L Moles of KNO3=0.6 M×0.4 L=0.24 mol\text{Moles of KNO}_3 = 0.6 \, \text{M} \times 0.4 \, \text{L} = 0.24 \, \text{mol}Moles of KNO3=0.6M×0.4L=0.24mol

Explanation: Multiply the molarity by the volume in liters to find the number of moles.

157. How many molecules are in 2.5 moles of H₂O?

Number of molecules=2.5 mol×6.022×1023 molecules/mol=1.506×1024 molecules\text{Number of molecules} = 2.5 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.506 \times 10^{24} \, \text{molecules}Number of molecules=2.5mol×6.022×1023molecules/mol=1.506×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

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Mole Concept (continued)

158. Calculate the number of grams of K₂Cr₂O₇ required to prepare 500 mL of a 0.2 M solution.

Moles of K2Cr2O7=0.2 M×0.5 L=0.1 mol\text{Moles of K}_2\text{Cr}_2\text{O}_7 = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{mol}Moles of K2Cr2O7=0.2M×0.5L=0.1mol Molecular weight of K2Cr2O7=2×39+2×52+7×16=294 g/mol\text{Molecular weight of K}_2\text{Cr}_2\text{O}_7 = 2 \times 39 + 2 \times 52 + 7 \times 16 = 294 \, \text{g/mol}Molecular weight of K2Cr2O7=2×39+2×52+7×16=294g/mol Mass of K2Cr2O7=0.1 mol×294 g/mol=29.4 g\text{Mass of K}_2\text{Cr}_2\text{O}_7 = 0.1 \, \text{mol} \times 294 \, \text{g/mol} = 29.4 \, \text{g}Mass of K2Cr2O7=0.1mol×294g/mol=29.4g

Explanation: Calculate the moles required and convert it to mass using the molecular weight.

159. Determine the number of moles in 200 mL of a 0.4 M HCl solution.

Volume of solution=200 mL=0.2 L\text{Volume of solution} = 200 \, \text{mL} = 0.2 \, \text{L}Volume of solution=200mL=0.2L Moles of HCl=0.4 M×0.2 L=0.08 mol\text{Moles of HCl} = 0.4 \, \text{M} \times 0.2 \, \text{L} = 0.08 \, \text{mol}Moles of HCl=0.4M×0.2L=0.08mol

Explanation: Multiply the molarity by the volume in liters to find the number of moles.

Determination of Formula of Compound (continued)

160. A compound is composed of 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 40 g
Hydrogen: 6.7 g
Oxygen: 53.3 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=3.33:6.7:3.33≈1:2:1\text{Ratio of C:H:O} = 3.33:6.7:3.33 \approx 1:2:1Ratio of C:H:O=3.33:6.7:3.33≈1:2:1

Step 4: Empirical formula:

Empirical formula=CH2O\text{Empirical formula} = \text{CH}_2\text{O}Empirical formula=CH2O

Explanation: Convert percentages to moles, simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

161. How many liters of 0.3 M H₂SO₄ are required to completely react with 2 moles of NaOH?

Balanced equation:

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Use stoichiometric ratio (1:2):

Moles of H2SO4=2 mol NaOH×1 mol H2SO42 mol NaOH=1 mol H2SO4\text{Moles of H}_2\text{SO}_4 = 2 \, \text{mol NaOH} \times \frac{1 \, \text{mol H}_2\text{SO}_4}{2 \, \text{mol NaOH}} = 1 \, \text{mol H}_2\text{SO}_4Moles of H2SO4=2mol NaOH×2mol NaOH1mol H2SO4=1mol H2SO4

Step 2: Calculate the volume of 0.3 M H₂SO₄ solution:

Volume of solution=moles of solutemolarity=1 mol0.3 M=3.33 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{1 \, \text{mol}}{0.3 \, \text{M}} = 3.33 \, \text{L}Volume of solution=molaritymoles of solute=0.3M1mol=3.33L

Explanation: Use the stoichiometric ratio and molarity to determine the volume of solution required.

Concentration Terms (continued)

162. Find the molarity of a solution prepared by dissolving 15 g of NaOH in 500 mL of solution.

Molecular weight of NaOH=23+16+1=40 g/mol\text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}Molecular weight of NaOH=23+16+1=40g/mol Moles of NaOH=15 g40 g/mol=0.375 mol\text{Moles of NaOH} = \frac{15 \, \text{g}}{40 \, \text{g/mol}} = 0.375 \, \text{mol}Moles of NaOH=40g/mol15g=0.375mol

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=0.375 mol0.5 L=0.75 M\text{Molarity} = \frac{0.375 \, \text{mol}}{0.5 \, \text{L}} = 0.75 \, \text{M}Molarity=0.5L0.375mol=0.75M

Explanation: Convert the mass of NaOH to moles and then find the molarity by dividing by the volume in liters.

Relation Between Molarity and Normality (continued)

163. Determine the normality of a 0.4 M H₃PO₄ solution.

Balanced equation:

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3 \text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=3\text{Number of H}^+ \text{ ions produced} = 3Number of H+ ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.4 M×3=1.2 N\text{Normality} = 0.4 \, \text{M} \times 3 = 1.2 \, \text{N}Normality=0.4M×3=1.2N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

164. How many grams of Na₂SO₄ are needed to prepare 1 liter of a 0.5 M solution?

Moles of Na2SO4=0.5 M×1 L=0.5 mol\text{Moles of Na}_2\text{SO}_4 = 0.5 \, \text{M} \times 1 \, \text{L} = 0.5 \, \text{mol}Moles of Na2SO4=0.5M×1L=0.5mol Molecular weight of Na2SO4=2×23+32+4×16=142 g/mol\text{Molecular weight of Na}_2\text{SO}_4 = 2 \times 23 + 32 + 4 \times 16 = 142 \, \text{g/mol}Molecular weight of Na2SO4=2×23+32+4×16=142g/mol Mass of Na2SO4=0.5 mol×142 g/mol=71 g\text{Mass of Na}_2\text{SO}_4 = 0.5 \, \text{mol} \times 142 \, \text{g/mol} = 71 \, \text{g}Mass of Na2SO4=0.5mol×142g/mol=71g

Explanation: Calculate the moles needed and convert to mass using the molecular weight.

165. Determine the number of molecules in 0.5 moles of H₂O.

Number of molecules=0.5 mol×6.022×1023 molecules/mol=3.011×1023 molecules\text{Number of molecules} = 0.5 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 3.011 \times 10^{23} \, \text{molecules}Number of molecules=0.5mol×6.022×1023molecules/mol=3.011×1023molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

166. Calculate the volume of 0.4 M NaCl required to obtain 0.6 moles of NaCl.

Volume of solution=moles of solutemolarity=0.6 mol0.4 M=1.5 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.6 \, \text{mol}}{0.4 \, \text{M}} = 1.5 \, \text{L}Volume of solution=molaritymoles of solute=0.4M0.6mol=1.5L

Explanation: Use the molarity formula to calculate the volume needed.

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Mole Concept (continued)

167. Calculate the number of grams of CuSO₄ required to prepare 250 mL of a 0.1 M solution.

Moles of CuSO4=0.1 M×0.25 L=0.025 mol\text{Moles of CuSO}_4 = 0.1 \, \text{M} \times 0.25 \, \text{L} = 0.025 \, \text{mol}Moles of CuSO4=0.1M×0.25L=0.025mol Molecular weight of CuSO4=63.5+32+4×16=159.5 g/mol\text{Molecular weight of CuSO}_4 = 63.5 + 32 + 4 \times 16 = 159.5 \, \text{g/mol}Molecular weight of CuSO4=63.5+32+4×16=159.5g/mol Mass of CuSO4=0.025 mol×159.5 g/mol=3.99 g\text{Mass of CuSO}_4 = 0.025 \, \text{mol} \times 159.5 \, \text{g/mol} = 3.99 \, \text{g}Mass of CuSO4=0.025mol×159.5g/mol=3.99g

Explanation: Convert the moles of CuSO₄ to grams using its molecular weight.

168. How many moles of NaOH are present in 300 mL of a 0.5 M solution?

Volume of solution=300 mL=0.3 L\text{Volume of solution} = 300 \, \text{mL} = 0.3 \, \text{L}Volume of solution=300mL=0.3L Moles of NaOH=0.5 M×0.3 L=0.15 mol\text{Moles of NaOH} = 0.5 \, \text{M} \times 0.3 \, \text{L} = 0.15 \, \text{mol}Moles of NaOH=0.5M×0.3L=0.15mol

Explanation: Multiply the molarity by the volume in liters to find the number of moles.

Determination of Formula of Compound (continued)

169. A compound contains 30% carbon, 10% hydrogen, and 60% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 30 g
Hydrogen: 10 g
Oxygen: 60 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=2.5:10:3.75≈2:8:3\text{Ratio of C:H:O} = 2.5:10:3.75 \approx 2:8:3Ratio of C:H:O=2.5:10:3.75≈2:8:3

Step 4: Empirical formula:

Empirical formula=C2H8O3\text{Empirical formula} = \text{C}_2\text{H}_8\text{O}_3Empirical formula=C2H8O3

Explanation: Convert percentages to moles, then simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

170. How many liters of 0.25 M Na₂SO₄ are required to react completely with 0.4 moles of HCl?

Balanced equation:

Na2SO4+2HCl→NaCl+H2SO4\text{Na}_2\text{SO}_4 + 2 \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{SO}_4Na2SO4+2HCl→NaCl+H2SO4

Step 1: Use stoichiometric ratio (1:2):

Moles of Na2SO4=0.4 mol HCl×1 mol Na2SO42 mol HCl=0.2 mol Na2SO4\text{Moles of Na}_2\text{SO}_4 = 0.4 \, \text{mol HCl} \times \frac{1 \, \text{mol Na}_2\text{SO}_4}{2 \, \text{mol HCl}} = 0.2 \, \text{mol Na}_2\text{SO}_4Moles of Na2SO4=0.4mol HCl×2mol HCl1mol Na2SO4=0.2mol Na2SO4

Step 2: Calculate the volume of 0.25 M Na₂SO₄ solution:

Volume of solution=moles of solutemolarity=0.2 mol0.25 M=0.8 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.2 \, \text{mol}}{0.25 \, \text{M}} = 0.8 \, \text{L}Volume of solution=molaritymoles of solute=0.25M0.2mol=0.8L

Explanation: Use the stoichiometric ratio and molarity to find the volume of solution required.

Concentration Terms (continued)

171. What is the molarity of a solution if 5 g of NaCl is dissolved in 200 mL of solution?

Molecular weight of NaCl=58.5 g/mol\text{Molecular weight of NaCl} = 58.5 \, \text{g/mol}Molecular weight of NaCl=58.5g/mol Moles of NaCl=5 g58.5 g/mol=0.085 mol\text{Moles of NaCl} = \frac{5 \, \text{g}}{58.5 \, \text{g/mol}} = 0.085 \, \text{mol}Moles of NaCl=58.5g/mol5g=0.085mol Volume of solution=200 mL=0.2 L\text{Volume of solution} = 200 \, \text{mL} = 0.2 \, \text{L}Volume of solution=200mL=0.2L Molarity=0.085 mol0.2 L=0.425 M\text{Molarity} = \frac{0.085 \, \text{mol}}{0.2 \, \text{L}} = 0.425 \, \text{M}Molarity=0.2L0.085mol=0.425M

Explanation: Convert mass to moles and then divide by the volume in liters to find the molarity.

Relation Between Molarity and Normality (continued)

172. Calculate the normality of a 0.6 M NaOH solution.

Balanced equation:

NaOH→Na++OH−\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-NaOH→Na++OH−

Step 1: Determine the number of equivalents:

Number of OH− ions produced=1\text{Number of OH}^- \text{ ions produced} = 1Number of OH− ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.6 M×1=0.6 N\text{Normality} = 0.6 \, \text{M} \times 1 = 0.6 \, \text{N}Normality=0.6M×1=0.6N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

173. How many moles of H₂SO₄ are in 500 mL of a 0.1 M solution?

Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Moles of H2SO4=0.1 M×0.5 L=0.05 mol\text{Moles of H}_2\text{SO}_4 = 0.1 \, \text{M} \times 0.5 \, \text{L} = 0.05 \, \text{mol}Moles of H2SO4=0.1M×0.5L=0.05mol

Explanation: Multiply the molarity by the volume in liters to find the number of moles.

174. Determine the mass of Na₂CO₃ required to prepare 250 mL of a 0.4 M solution.

Moles of Na2CO3=0.4 M×0.25 L=0.1 mol\text{Moles of Na}_2\text{CO}_3 = 0.4 \, \text{M} \times 0.25 \, \text{L} = 0.1 \, \text{mol}Moles of Na2CO3=0.4M×0.25L=0.1mol Molecular weight of Na2CO3=2×23+12+3×16=106 g/mol\text{Molecular weight of Na}_2\text{CO}_3 = 2 \times 23 + 12 + 3 \times 16 = 106 \, \text{g/mol}Molecular weight of Na2CO3=2×23+12+3×16=106g/mol Mass of Na2CO3=0.1 mol×106 g/mol=10.6 g\text{Mass of Na}_2\text{CO}_3 = 0.1 \, \text{mol} \times 106 \, \text{g/mol} = 10.6 \, \text{g}Mass of Na2CO3=0.1mol×106g/mol=10.6g

Explanation: Calculate the moles needed and convert to mass using the molecular weight.

175. How many molecules are in 0.25 moles of CO₂?

Number of molecules=0.25 mol×6.022×1023 molecules/mol=1.506×1023 molecules\text{Number of molecules} = 0.25 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.506 \times 10^{23} \, \text{molecules}Number of molecules=0.25mol×6.022×1023molecules/mol=1.506×1023molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

176. Calculate the volume of 0.6 M H₂SO₄ required to obtain 0.4 moles of H₂SO₄.

Volume of solution=moles of solutemolarity=0.4 mol0.6 M=0.667 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.4 \, \text{mol}}{0.6 \, \text{M}} = 0.667 \, \text{L}Volume of solution=molaritymoles of solute=0.6M0.4mol=0.667L

Explanation: Use the molarity formula to determine the volume needed.

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Mole Concept (continued)

177. How many molecules are in 5 g of H₂O?

Step 1: Convert grams to moles:

Molecular weight of H2O=2×1+16=18 g/mol\text{Molecular weight of H}_2\text{O} = 2 \times 1 + 16 = 18 \, \text{g/mol}Molecular weight of H2O=2×1+16=18g/mol Moles of H2O=5 g18 g/mol=0.278 mol\text{Moles of H}_2\text{O} = \frac{5 \, \text{g}}{18 \, \text{g/mol}} = 0.278 \, \text{mol}Moles of H2O=18g/mol5g=0.278mol

Step 2: Calculate the number of molecules:

Number of molecules=0.278 mol×6.022×1023 molecules/mol=1.674×1023 molecules\text{Number of molecules} = 0.278 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.674 \times 10^{23} \, \text{molecules}Number of molecules=0.278mol×6.022×1023molecules/mol=1.674×1023molecules

Explanation: Convert the mass to moles and then multiply by Avogadro's number to find the number of molecules.

178. Determine the volume of 1.5 M NaOH solution needed to get 0.75 moles of NaOH.

Volume of solution=moles of solutemolarity=0.75 mol1.5 M=0.5 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.75 \, \text{mol}}{1.5 \, \text{M}} = 0.5 \, \text{L}Volume of solution=molaritymoles of solute=1.5M0.75mol=0.5L

Explanation: Use the molarity formula to find the volume required.

Determination of Formula of Compound (continued)

179. A compound has the following percentage composition: 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Carbon: 40 g
Hydrogen: 6.7 g
Oxygen: 53.3 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of C:H:O=3.33:6.7:3.33≈1:2:1\text{Ratio of C:H:O} = 3.33:6.7:3.33 \approx 1:2:1Ratio of C:H:O=3.33:6.7:3.33≈1:2:1

Step 4: Empirical formula:

Empirical formula=CH2O\text{Empirical formula} = \text{CH}_2\text{O}Empirical formula=CH2O

Explanation: Convert percentages to moles, then simplify the mole ratio to determine the empirical formula.

Stoichiometric Calculations (continued)

180. How many grams of NaOH are required to react with 0.4 moles of H₂SO₄?

Balanced equation:

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Calculate moles of NaOH needed:

Moles of NaOH=0.4 mol H2SO4×2=0.8 mol NaOH\text{Moles of NaOH} = 0.4 \, \text{mol H}_2\text{SO}_4 \times 2 = 0.8 \, \text{mol NaOH}Moles of NaOH=0.4mol H2SO4×2=0.8mol NaOH

Step 2: Convert moles to grams:

Molecular weight of NaOH=23+16+1=40 g/mol\text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}Molecular weight of NaOH=23+16+1=40g/mol Mass of NaOH=0.8 mol×40 g/mol=32 g\text{Mass of NaOH} = 0.8 \, \text{mol} \times 40 \, \text{g/mol} = 32 \, \text{g}Mass of NaOH=0.8mol×40g/mol=32g

Explanation: Use the stoichiometric ratio to find moles and then convert to mass.

Concentration Terms (continued)

181. Find the molarity of a solution prepared by dissolving 20 g of KCl in 500 mL of solution.

Molecular weight of KCl=39+35.5=74.5 g/mol\text{Molecular weight of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol}Molecular weight of KCl=39+35.5=74.5g/mol Moles of KCl=20 g74.5 g/mol=0.268 mol\text{Moles of KCl} = \frac{20 \, \text{g}}{74.5 \, \text{g/mol}} = 0.268 \, \text{mol}Moles of KCl=74.5g/mol20g=0.268mol Volume of solution=500 mL=0.5 L\text{Volume of solution} = 500 \, \text{mL} = 0.5 \, \text{L}Volume of solution=500mL=0.5L Molarity=0.268 mol0.5 L=0.536 M\text{Molarity} = \frac{0.268 \, \text{mol}}{0.5 \, \text{L}} = 0.536 \, \text{M}Molarity=0.5L0.268mol=0.536M

Explanation: Convert mass to moles and then divide by the volume in liters to find the molarity.

Relation Between Molarity and Normality (continued)

182. Calculate the normality of a 0.2 M HCl solution.

Balanced equation:

HCl→H++Cl−\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-HCl→H++Cl−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=1\text{Number of H}^+ \text{ ions produced} = 1Number of H+ ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.2 M×1=0.2 N\text{Normality} = 0.2 \, \text{M} \times 1 = 0.2 \, \text{N}Normality=0.2M×1=0.2N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

183. How many grams of NaCl are needed to prepare 1 L of a 0.2 M solution?

Moles of NaCl=0.2 M×1 L=0.2 mol\text{Moles of NaCl} = 0.2 \, \text{M} \times 1 \, \text{L} = 0.2 \, \text{mol}Moles of NaCl=0.2M×1L=0.2mol Molecular weight of NaCl=58.5 g/mol\text{Molecular weight of NaCl} = 58.5 \, \text{g/mol}Molecular weight of NaCl=58.5g/mol Mass of NaCl=0.2 mol×58.5 g/mol=11.7 g\text{Mass of NaCl} = 0.2 \, \text{mol} \times 58.5 \, \text{g/mol} = 11.7 \, \text{g}Mass of NaCl=0.2mol×58.5g/mol=11.7g

Explanation: Calculate the moles needed and convert to mass using the molecular weight.

184. Determine the volume of 0.4 M NaCl solution needed to obtain 0.2 moles of NaCl.

Volume of solution=moles of solutemolarity=0.2 mol0.4 M=0.5 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.2 \, \text{mol}}{0.4 \, \text{M}} = 0.5 \, \text{L}Volume of solution=molaritymoles of solute=0.4M0.2mol=0.5L

Explanation: Use the molarity formula to determine the volume needed.

185. How many molecules are present in 2 moles of CO₂?

Number of molecules=2 mol×6.022×1023 molecules/mol=1.204×1024 molecules\text{Number of molecules} = 2 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.204 \times 10^{24} \, \text{molecules}Number of molecules=2mol×6.022×1023molecules/mol=1.204×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

186. Calculate the volume of 0.5 M H₂SO₄ solution needed to get 0.25 moles of H₂SO₄.

Volume of solution=moles of solutemolarity=0.25 mol0.5 M=0.5 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.25 \, \text{mol}}{0.5 \, \text{M}} = 0.5 \, \text{L}Volume of solution=molaritymoles of solute=0.5M0.25mol=0.5L

Explanation: Use the molarity formula to determine the volume required.

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Mole Concept (continued)

187. How many grams of glucose (C₆H₁₂O₆) are there in 0.4 moles of the substance?

Explanation: Multiply the moles by the molecular weight to get the mass.

188. Determine the number of moles in 50 g of KNO₃.

Molecular weight of KNO3=39+14+3×16=101 g/mol\text{Molecular weight of KNO}_3 = 39 + 14 + 3 \times 16 = 101 \, \text{g/mol}Molecular weight of KNO3=39+14+3×16=101g/mol Moles of KNO3=50 g101 g/mol=0.495 mol\text{Moles of KNO}_3 = \frac{50 \, \text{g}}{101 \, \text{g/mol}} = 0.495 \, \text{mol}Moles of KNO3=101g/mol50g=0.495mol

Explanation: Divide the mass by the molecular weight to find the number of moles.

Determination of Formula of Compound (continued)

189. A compound contains 20% sulfur and 80% oxygen by mass. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Sulfur: 20 g
Oxygen: 80 g

Step 2: Convert mass to moles:

Moles of S=20 g32 g/mol=0.625 mol\text{Moles of S} = \frac{20 \, \text{g}}{32 \, \text{g/mol}} = 0.625 \, \text{mol}Moles of S=32g/mol20g=0.625mol Moles of O=80 g16 g/mol=5 mol\text{Moles of O} = \frac{80 \, \text{g}}{16 \, \text{g/mol}} = 5 \, \text{mol}Moles of O=16g/mol80g=5mol

Step 3: Determine the simplest mole ratio:

Ratio of S:O=0.625:5≈1:8\text{Ratio of S:O} = 0.625:5 \approx 1:8Ratio of S:O=0.625:5≈1:8

Step 4: Empirical formula:

Empirical formula=SO8\text{Empirical formula} = \text{SO}_8Empirical formula=SO8

Explanation: Convert percentages to moles, simplify the mole ratio to obtain the empirical formula.

Stoichiometric Calculations (continued)

190. Calculate the volume of 0.2 M H₂SO₄ required to neutralize 0.4 moles of NaOH.

Balanced equation:

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}H2SO4+2NaOH→Na2SO4+2H2O

Step 1: Use stoichiometric ratio (1:2):

Moles of H2SO4=0.4 mol NaOH2=0.2 mol H2SO4\text{Moles of H}_2\text{SO}_4 = \frac{0.4 \, \text{mol NaOH}}{2} = 0.2 \, \text{mol H}_2\text{SO}_4Moles of H2SO4=20.4mol NaOH=0.2mol H2SO4

Step 2: Calculate the volume of 0.2 M H₂SO₄ solution:

Volume of solution=0.2 mol0.2 M=1 L\text{Volume of solution} = \frac{0.2 \, \text{mol}}{0.2 \, \text{M}} = 1 \, \text{L}Volume of solution=0.2M0.2mol=1L

Explanation: Use the stoichiometric ratio to determine the moles required and then calculate the volume.

Concentration Terms (continued)

191. What is the molarity of a solution if 10 g of Na₂SO₄ is dissolved in 250 mL of solution?

Molecular weight of Na2SO4=2×23+32+4×16=142 g/mol\text{Molecular weight of Na}_2\text{SO}_4 = 2 \times 23 + 32 + 4 \times 16 = 142 \, \text{g/mol}Molecular weight of Na2SO4=2×23+32+4×16=142g/mol Moles of Na2SO4=10 g142 g/mol=0.0704 mol\text{Moles of Na}_2\text{SO}_4 = \frac{10 \, \text{g}}{142 \, \text{g/mol}} = 0.0704 \, \text{mol}Moles of Na2SO4=142g/mol10g=0.0704mol Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=0.0704 mol0.25 L=0.2816 M\text{Molarity} = \frac{0.0704 \, \text{mol}}{0.25 \, \text{L}} = 0.2816 \, \text{M}Molarity=0.25L0.0704mol=0.2816M

Explanation: Convert the mass to moles and then divide by the volume in liters to get the molarity.

Relation Between Molarity and Normality (continued)

192. Determine the normality of a 0.5 M HNO₃ solution.

Balanced equation:

HNO3→H++NO3−\text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^-HNO3→H++NO3−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=1\text{Number of H}^+ \text{ ions produced} = 1Number of H+ ions produced=1

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.5 M×1=0.5 N\text{Normality} = 0.5 \, \text{M} \times 1 = 0.5 \, \text{N}Normality=0.5M×1=0.5N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

193. How many grams of NaHCO₃ are needed to prepare 500 mL of a 0.2 M solution?

Moles of NaHCO3=0.2 M×0.5 L=0.1 mol\text{Moles of NaHCO}_3 = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{mol}Moles of NaHCO3=0.2M×0.5L=0.1mol Molecular weight of NaHCO3=23+1+12+3×16=84 g/mol\text{Molecular weight of NaHCO}_3 = 23 + 1 + 12 + 3 \times 16 = 84 \, \text{g/mol}Molecular weight of NaHCO3=23+1+12+3×16=84g/mol Mass of NaHCO3=0.1 mol×84 g/mol=8.4 g\text{Mass of NaHCO}_3 = 0.1 \, \text{mol} \times 84 \, \text{g/mol} = 8.4 \, \text{g}Mass of NaHCO3=0.1mol×84g/mol=8.4g

Explanation: Calculate the moles needed and convert to mass using the molecular weight.

194. Calculate the volume of 0.3 M KCl solution required to obtain 0.6 moles of KCl.

Volume of solution=moles of solutemolarity=0.6 mol0.3 M=2 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.6 \, \text{mol}}{0.3 \, \text{M}} = 2 \, \text{L}Volume of solution=molaritymoles of solute=0.3M0.6mol=2L

Explanation: Use the molarity formula to determine the volume needed.

195. Determine the number of moles in 100 mL of 0.25 M solution of H₂SO₄.

Volume of solution=100 mL=0.1 L\text{Volume of solution} = 100 \, \text{mL} = 0.1 \, \text{L}Volume of solution=100mL=0.1L Moles of H2SO4=0.25 M×0.1 L=0.025 mol\text{Moles of H}_2\text{SO}_4 = 0.25 \, \text{M} \times 0.1 \, \text{L} = 0.025 \, \text{mol}Moles of H2SO4=0.25M×0.1L=0.025mol

Explanation: Multiply the molarity by the volume in liters to find the number of moles.

196. How many molecules are in 3 moles of Na₂SO₄?

Number of molecules=3 mol×6.022×1023 molecules/mol=1.806×1024 molecules\text{Number of molecules} = 3 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.806 \times 10^{24} \, \text{molecules}Number of molecules=3mol×6.022×1023molecules/mol=1.806×1024molecules

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

197. Calculate the normality of a solution that is 0.4 M H₂SO₄.

Balanced equation:

H2SO4→2H++SO42−\text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-}H2SO4→2H++SO42−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=2\text{Number of H}^+ \text{ ions produced} = 2Number of H+ ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.4 M×2=0.8 N\text{Normality} = 0.4 \, \text{M} \times 2 = 0.8 \, \text{N}Normality=0.4M×2=0.8N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

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Mole Concept (continued)

198. How many moles of CO₂ are there in 88 grams of CO₂?

Molecular weight of CO2=12+2×16=44 g/mol\text{Molecular weight of CO}_2 = 12 + 2 \times 16 = 44 \, \text{g/mol}Molecular weight of CO2=12+2×16=44g/mol Moles of CO2=88 g44 g/mol=2 mol\text{Moles of CO}_2 = \frac{88 \, \text{g}}{44 \, \text{g/mol}} = 2 \, \text{mol}Moles of CO2=44g/mol88g=2mol

Explanation: Divide the mass by the molecular weight to find the number of moles.

199. Calculate the mass of 0.75 moles of NaCl.

Molecular weight of NaCl=23+35.5=58.5 g/mol\text{Molecular weight of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}Molecular weight of NaCl=23+35.5=58.5g/mol Mass of NaCl=0.75 mol×58.5 g/mol=43.875 g\text{Mass of NaCl} = 0.75 \, \text{mol} \times 58.5 \, \text{g/mol} = 43.875 \, \text{g}Mass of NaCl=0.75mol×58.5g/mol=43.875g

Explanation: Multiply the number of moles by the molecular weight to get the mass.

Determination of Formula of Compound (continued)

200. A compound is composed of 30% nitrogen and 70% oxygen. Determine its empirical formula.

Step 1: Assume 100 g of the compound:

Nitrogen: 30 g
Oxygen: 70 g

Step 2: Convert mass to moles:

Step 3: Determine the simplest mole ratio:

Ratio of N:O=2.14:4.38≈1:2\text{Ratio of N:O} = 2.14:4.38 \approx 1:2Ratio of N:O=2.14:4.38≈1:2

Step 4: Empirical formula:

Empirical formula=NO2\text{Empirical formula} = \text{NO}_2Empirical formula=NO2

Explanation: Convert percentages to moles, simplify the mole ratio to determine the empirical formula.

Stoichiometric Calculations (continued)

201. How many grams of KCl are needed to react with 0.5 moles of AgNO₃ to form AgCl and KNO₃?

Balanced equation:

AgNO3+KCl→AgCl+KNO3\text{AgNO}_3 + \text{KCl} \rightarrow \text{AgCl} + \text{KNO}_3AgNO3+KCl→AgCl+KNO3

Step 1: Use stoichiometric ratio (1:1):

Moles of KCl needed=0.5 mol\text{Moles of KCl needed} = 0.5 \, \text{mol}Moles of KCl needed=0.5mol

Step 2: Convert moles to grams:

Molecular weight of KCl=39+35.5=74.5 g/mol\text{Molecular weight of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol}Molecular weight of KCl=39+35.5=74.5g/mol Mass of KCl=0.5 mol×74.5 g/mol=37.25 g\text{Mass of KCl} = 0.5 \, \text{mol} \times 74.5 \, \text{g/mol} = 37.25 \, \text{g}Mass of KCl=0.5mol×74.5g/mol=37.25g

Explanation: Use the stoichiometric ratio to find moles and then convert to mass using the molecular weight.

Concentration Terms (continued)

202. Find the molarity of a solution if 12 g of HCl is dissolved in 250 mL of solution.

Molecular weight of HCl=1+35.5=36.5 g/mol\text{Molecular weight of HCl} = 1 + 35.5 = 36.5 \, \text{g/mol}Molecular weight of HCl=1+35.5=36.5g/mol Moles of HCl=12 g36.5 g/mol=0.328 mol\text{Moles of HCl} = \frac{12 \, \text{g}}{36.5 \, \text{g/mol}} = 0.328 \, \text{mol}Moles of HCl=36.5g/mol12g=0.328mol Volume of solution=250 mL=0.25 L\text{Volume of solution} = 250 \, \text{mL} = 0.25 \, \text{L}Volume of solution=250mL=0.25L Molarity=0.328 mol0.25 L=1.312 M\text{Molarity} = \frac{0.328 \, \text{mol}}{0.25 \, \text{L}} = 1.312 \, \text{M}Molarity=0.25L0.328mol=1.312M

Explanation: Convert the mass to moles and then divide by the volume in liters to find the molarity.

Relation Between Molarity and Normality (continued)

203. Determine the normality of a 0.3 M H₃PO₄ solution.

Balanced equation:

H3PO4→3H++PO43−\text{H}_3\text{PO}_4 \rightarrow 3 \text{H}^+ + \text{PO}_4^{3-}H3PO4→3H++PO43−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=3\text{Number of H}^+ \text{ ions produced} = 3Number of H+ ions produced=3

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.3 M×3=0.9 N\text{Normality} = 0.3 \, \text{M} \times 3 = 0.9 \, \text{N}Normality=0.3M×3=0.9N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

Additional Numerical Problems with Explanations (continued)

204. How many grams of Na₂SO₄ are needed to prepare 500 mL of a 0.4 M solution?

Moles of Na2SO4=0.4 M×0.5 L=0.2 mol\text{Moles of Na}_2\text{SO}_4 = 0.4 \, \text{M} \times 0.5 \, \text{L} = 0.2 \, \text{mol}Moles of Na2SO4=0.4M×0.5L=0.2mol Molecular weight of Na2SO4=2×23+32+4×16=142 g/mol\text{Molecular weight of Na}_2\text{SO}_4 = 2 \times 23 + 32 + 4 \times 16 = 142 \, \text{g/mol}Molecular weight of Na2SO4=2×23+32+4×16=142g/mol Mass of Na2SO4=0.2 mol×142 g/mol=28.4 g\text{Mass of Na}_2\text{SO}_4 = 0.2 \, \text{mol} \times 142 \, \text{g/mol} = 28.4 \, \text{g}Mass of Na2SO4=0.2mol×142g/mol=28.4g

Explanation: Calculate the moles needed and convert to mass using the molecular weight.

205. Determine the volume of 0.25 M HNO₃ solution required to obtain 0.75 moles of HNO₃.

Volume of solution=moles of solutemolarity=0.75 mol0.25 M=3 L\text{Volume of solution} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.75 \, \text{mol}}{0.25 \, \text{M}} = 3 \, \text{L}Volume of solution=molaritymoles of solute=0.25M0.75mol=3L

Explanation: Use the molarity formula to determine the volume needed.

206. How many molecules are there in 4 moles of Na₂CO₃?

Explanation: Multiply the number of moles by Avogadro's number to find the number of molecules.

207. Calculate the normality of a 0.4 M H₂SO₄ solution.

Balanced equation:

H2SO4→2H++SO42−\text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-}H2SO4→2H++SO42−

Step 1: Determine the number of equivalents:

Number of H+ ions produced=2\text{Number of H}^+ \text{ ions produced} = 2Number of H+ ions produced=2

Step 2: Use the relationship between molarity and normality:

Normality=Molarity×number of equivalents\text{Normality} = \text{Molarity} \times \text{number of equivalents}Normality=Molarity×number of equivalents Normality=0.4 M×2=0.8 N\text{Normality} = 0.4 \, \text{M} \times 2 = 0.8 \, \text{N}Normality=0.4M×2=0.8N

Explanation: Multiply the molarity by the number of equivalents to find the normality.

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UNIT 1Mole Concept

Some Basic Concepts of ChemistryMole Concept, Determination of Formula of Compound, Stoichiometric Calculations, Concentration Terms, Relation Between Molarity and Normality

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