Calculate the molalities of some commercial reagents from the following

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