Question

Given that the vapor pressure of water is 17.54 torr at 20 °c, calculate the vapor-pressure lowering of aqueous solutions that are 1.80 m in (a) sucrose, c12h22o11, and (b) sodium chloride. assume 100% dissociation for electrolytes.

Answer 1

1) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) = 1000 g ÷ 1000 g/kg = 1 kg.
n(sucrose - C₁₂O₂₂O₁₁) = b(solution) · m(H₂O).
n(C₁₂O₂₂O₁₁) = 1,8 mol/kg · 1 kg.
n(C₁₂O₂₂O₁₁) = 1,8 mol.
n(H₂O) = 1000 g ÷ 18 g/mol.
n(H₂O) = 55,55 mol.
Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol) = 0,968.
Raoult's Law: p(solution) = mole fraction of solvent · p(solvent).
p(solution) = 0,968 · 17,54 torr = 16,99 torr.
Δp = 17,54 torr - 16,87 torr = 0,55 torr.

2) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) = 1000 g ÷ 1000 g/kg = 1 kg.
n(NaCl) = b(solution) · m(H₂O).
n(NaCl) = 1,8 mol/kg · 1 kg.
n(NaCl) = 1,8 mol.
n(H₂O) = 1000 g ÷ 18 g/mol.
n(H₂O) = 55,55 mol.
i(NaCl) = 2; Van 't Hoff factor. Because dissociate on one cation and one anions.

Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol · 2) = 0,94.

Raoult's Law: p(solution) = mole fraction of solvent · p(solvent)
p(solution) = 0,94 · 17,54 torr = 16,47 torr.
Δp = 17,54 torr - 16,47 torr = 1,06 torr.