Question

An aqueous solution is formed by dissolving 8.00 mol of a non-electrolyte in 1.00 kg of water. If the vapor pressure of pure water is 23.8 mm Hg at 25/C, the vapor pressure of the solution will be:

1. 19.4 mm Hg

2. 23.8 mm Hg

3. 5.5 mm Hg

4. 26.8 mm Hg

5. 20.8 mm Hg

Answer 1

Ans: vapor pressure of the solution = 20.804 mm Hg

Mass of water = 1.0 kg = 1000 g

Molar mass of water = 18 g/mol

No. of moles of water, n1 = 1000/18 = 55.55 mol

No. of moles of non-electrolyte, n2 = 8.0 mol

Mole fraction of water in solution, X1 = n1/(n1+n2) = 55.55/(8.0 + 55.55) = 0.8741

Mole fraction of non-electrolyte in the solution, X2 = n2/(n1+n1) = 8.0/(8.0 + 55.55) = 0.1259

If, P^o (23.8 mm Hg) is the vapor pressure of pure water and P is the vapor pressure of solution, then

According to Raoult’s law, P = P^oX₁ = 23.8 x 0.8741 = 20.804 mm Hg