What is the vapour pressure of a solution by mixing 35g of solid Na2SO4 with 175ml of water at 25?

Molar mass of Na₂SO₄ = (23.0×2 + 32.1 + 16.0×4) g/mol = 142.1 g/mol
No. of moles of Na₂SO₄ added = (35 g) / (142.1 g/mol) = 0.246 mol
1 mole of Na₂SO₄ contains 3 moles of ions (2 moles of Na⁺ ions and 1 mole of SO₄²⁻ ions).
No. of moles of ions in the solution = (0.246 mol) × 2 = 0.738 mol

Density of water = 1.00 g/ml
Mass of water in the solution = (1.00 g/ml) × (175 ml) = 175 g
Molar mass of water (H₂O) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
No. of moles of water in the solution = (175 g) / (18.0 g/mol) = 9.72 mol
Molar fraction of water in the solution, X(H₂O) = 9.72 / (9.72 + 0.738) = 0.929

Vapor pressure of pure water at 25°C, P°(H₂O) = 23.8 torr

By Raoult s law :
Vapor pressure of the solution at 25°C, P(Na₂SO₄) = P°(H₂O) × X(H₂O) = (23.8 torr) × 0.929 = 22.1 torr