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.

Answer is (d)a brainliest would be appreciated

(no3)⁻n has oxidation number x,o has oxidation number -2.(x*1 -2*3) charge of the nitrate ion, and we know that charge of the nitrate ion =-1.so,(x*1 -2*3) =-1x-6=-1x=+5answer a. +5.