Question

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?

Answer 1

Answer:

[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