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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.

Respuesta :

Dejanras
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.
ogorwyne

The mole of water at 1000 grams

= [tex]\frac{1000}{18} = 55.55[/tex]

Molarity of the solution = 2.2

The amount of water = moles of solvent

Mole fraction

[tex]\frac{55.55}{55.55+2} =0.9619\\\\[/tex]

From Raoults law

Vapor pressure = mole fraction * vapor pressure

= 17.54*0.9619

= 16.87 torr

Lowered vapor pressure

17.54 - 16.87

= 0.67 torr

2. Mole fraction of solvent

[tex]\frac{55.55}{55.55+2.2*2} =0.93[/tex]

0.93*17.54 = 16.31

17.54-16.31 = 1.23

The lowered vapor with NACl = 1.23

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