Given that the vapor pressure of water is 17.54 torr at 20 °c, calculate the vapor-pressure lowering of aqueous solutions that are 2.20 m in (a) sucrose, c12h22o11, and (b) aluminum chloride. assume 100% dissociation for electrolytes.

Question

Macie Cobb

Chemistry

19 April, 14:37

Given that the vapor pressure of water is 17.54 torr at 20 °c, calculate the vapor-pressure lowering of aqueous solutions that are 2.20 m in (a) sucrose, c12h22o11, and (b) aluminum chloride. assume 100% dissociation for electrolytes.

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Kaia Atkinson · Answer 1

1) b (solution) = 2,20 m = 2,2 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) = 2,2 mol/kg · 1 kg.

n (C₁₂O₂₂O₁₁) = 2,2 mol.

n (H₂O) = 1000 g : 18 g/mol.

n (H₂O) = 55,55 mol.

Mole fraktion of solvent = 55,55 mol : (55,55 mol + 2,2 mol) = 0,962.

Raoult's Law: p (solution) = mole fraction of solvent · p (solvent).

p (solution) = 0,962 · 17,54 torr = 16,87 torr.

Δp = 17,54 torr - 16,87 torr = 0,67 torr.

2) b (solution) = 2,20 m = 2,2 mol/kg ...

If we use 1000 g of water to make solution:

m (H₂O) = 1000 g : 1000 g/kg = 1 kg.

n (AlCl₃) = b (solution) · m (H₂O).

n (AlCl₃) = 2,2 mol/kg · 1 kg.

n (AlCl₃) = 2,2 mol.

n (H₂O) = 1000 g : 18 g/mol.

n (H₂O) = 55,55 mol.

i (AlCl₃) = 4; Van 't Hoff factor. Because dissociate on one cation and three anions. Mole fraktion of solvent = 55,55 mol : (55,55 mol + 2,2 mol · 4) = 0,863.

Raoult's Law: p (solution) = mole fraction of solvent · p (solvent)

p (solution) = 0,863 · 17,54 torr = 15,16 torr.

Δp = 17,54 torr - 15,16 torr = 1,38 torr.