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25 April, 01:47

Common commercial acids and bases are aqueous solutions with the following properties: Density Mass Percent of Solute Hydrochloric acid 1.19 38 Nitric acid 1.42 70. Sulfuric acid 1.84 95 Acetic acid 1.05 99 Ammonia 0.90 28 Calculate the molarity, molality, and mole fraction of each of the preceding reagents?

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  1. 25 April, 01:56
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    [HCl] : 12.3 M; [HCl] : 16.7 m; [HCl] in mole fraction = 0.23

    [HNO₃]: 15.7 M; [HNO₃]: 30 m; [HNO₃] in mole fraction = 0.39

    [H₂SO₄]: 17.8 M; [H₂SO₄]: 193.8 m; [H₂SO₄] in mole fraction = 0.78

    [CH₃COOH]: 17.3 M; [CH₃COOH]: 1650 m; [CH₃COOH] in mole fraction: 0.96

    [NH₃]: 14.7 M; [NH₃]: 22.7 m; [NH₃] in mole fraction = 0.29

    Explanation:

    Percent by mass means X g of solute contained in 100 g of solution

    Mass of solvent: 100 g - X % by mass

    Solution density = Solution mass / Solution volume

    Solution Volume = Solution mass / Solution density

    Molality = mol of solute / kg of solvent

    Molarity = mol of solute / L of solution

    Mole fraction = Moles of solute / Total moles

    Total moles = moles of solute + moles of solvent

    HCl → 1.19 g/mL; 38%

    Mass of solvent: 100 g - 38 g = 62 g

    Moles of solvent → 62 g. 1 mol / 18g = 3.44 moles

    Mass of solute → 38 g; Moles of solute → 38 g. 1mol/36.45 g = 1.04 moles

    Total moles = 3.44 mol + 1.04 mol = 4.48 moles

    Mole fraction = 1.04 / 4.48 = 0.23

    Mass of solvent → from g to kg → 62 g. 1kg/1000g = 0.062kg

    Molality → 1.04 mol / 0.062 kg = 16.7 mol/kg → 16.7 m

    Solution volume = 100 g / 1.19 g/mL → 84.03 mL

    Solution volume → from mL to L → 84.03 mL. 1L / 1000mL = 0.0840L

    Molarity → 1.04 mol / 0.0840L = 12.3 M

    HNO₃ → 1.42 g/mL; 70%

    Mass of solvent: 100 g - 70 g = 30 g

    Moles of solvent → 30 g. 1 mol / 18g = 1.67 moles

    Mass of solute → 70 g; Moles of solute → 70 g. 1mol / 63 g = 1.11 moles

    Total moles = 1.67 mol + 1.11 mol = 2.78 moles

    Mole fraction = 1.11 / 2.78 = 0.39

    Mass of solvent → from g to kg → 30 g. 1kg/1000g = 0.030kg

    Molality → 1.11 mol / 0.030 kg → 30 m

    Solution volume = 100 g / 1.42 g/mL → 70.4mL

    Solution volume → from mL to L → 70.4 mL. 1L / 1000mL = 0.0704L

    Molarity → 1.11 mol / 0.0704L = 15.7 M

    H₂SO₄ → 1.84 g/mL; 95 %

    Mass of solvent: 100 g - 95 g = 5 g

    Moles of solvent → 5 g. 1 mol / 18g = 0.277 moles

    Mass of solute → 95 g; Moles of solute → 95 g. 1 mol / 98 g = 0.969 moles

    Total moles = 0.277 mol + 0.969 mol = 1.246 moles

    Mole fraction = 0.969 / 1.246 = 0.78

    Mass of solvent → from g to kg → 5 g. 1kg/1000g = 0.005 kg

    Molality → 0.969 mol / 0.005 kg → 193.8 m

    Solution volume = 100 g / 1.84 g/mL → 54.3 mL

    Solution volume → from mL to L → 54.3 mL. 1L / 1000mL = 0.0543 L

    Molarity → 0.969 mol / 0.0543L = 17.8 M

    CH₃COOH → 1.05 g/mL; 99 %

    Mass of solvent: 100 g - 99 g = 1 g

    Moles of solvent → 1 g. 1 mol / 18g = 0.055 moles

    Mass of solute → 99 g; Moles of solute → 99 g. 1 mol / 60 g = 1.65 moles

    Total moles = 0.055 mol + 1.65 mol = 1.705 moles

    Mole fraction = 1.65 / 1.705 = 0.96

    Mass of solvent → from g to kg → 1 g. 1kg/1000 g = 0.001 kg

    Molality → 1.65 mol / 0.001 kg → 1650 m

    Solution volume = 100 g / 1.05 g/mL → 95.2mL

    Solution volume → from mL to L → 95.2 mL. 1L / 1000mL = 0.0952L

    Molarity → 1.65 mol / 0.0952 L = 17.3 M

    NH₃ → 0.90 g/mL; 28 %

    Mass of solvent: 100 g - 28 g = 72 g

    Moles of solvent → 72 g. 1 mol / 18g = 4 moles

    Mass of solute → 28 g; Moles of solute → 28 g. 1mol / 17 g = 1.64 moles

    Total moles = 4 mol + 1.64 mol = 5.64 moles

    Mole fraction = 1.64 / 5.64 = 0.29

    Mass of solvent → from g to kg → 72 g. 1kg/1000g = 0.072kg

    Molality → 1.64 mol / 0.072 kg → 22.7 m

    Solution volume = 100 g / 0.90 g/mL → 111.1mL

    Solution volume → from mL to L → 111.1 mL. 1L / 1000mL = 0.111L

    Molarity → 1.64 mol / 0.111L = 14.7 M
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