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.

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.

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.

Ibelieve the answer is d. titaniumhope my answer has you : )