The relative lowering of vapour pressure produced by dissolving 18 g of urea (Molar mass = 60 g mol ⁻¹ ) in 100 g of water is

Step-by-step solution

Moles of urea, n₂ = 18 60 = 0.3 mol Moles of water, n₁ = 100 18 = 5.55 mol According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute. P P^ = ₂ = n₂ n₁ + n₂ ₂ = 0.3 5.55 + 0.3 = 0.3 5.85 = 0.0512 Rounding to three decimal places, we get 0.051 . Answer: 0.051