Mass of Urea (NH₂ CONH₂) required to be dissolved in 1000 g of water in order to reduce the vapour pressure of water by 25 % is _________ g. (Nearest integer)
Given: Molar mass of N, C, O and H are 14,12,16 and 1 g mol⁻¹ respectively

Solution

Chemistry ma’am · 4 steps

A 25% drop is far from dilute, so use the full mole fraction of urea, not the n/N shortcut.

1. Raoult's law. The relative lowering of vapour pressure equals the mole fraction of the solute. A 25% drop means xᵤᵣₑₐ = 0.25.

   (p° - p)/p° = n/(n + N) = 0.25

2. Moles of water. Molar mass of water is 18 g mol⁻¹, so N = 1000/18 = 55.56 mol. From n/(n+N) = 1/4 you get 3n = N, so n = 55.56/3 = 18.52 mol.
3. Convert to mass. Molar mass of urea is 2(14) + 4(1) + 12 + 16 = 60 g mol⁻¹. Mass = 18.52 × 60 = 1111.1 g.
4. So. The mass of urea needed is about 1111 g.

The common slip: 833. Using the dilute shortcut n/N = 0.25 gives n = 13.89 mol and a mass of 833 g, not 1111.

The idea to keep. The relative lowering of vapour pressure equals n/(n+N). Use n/N only for very dilute solutions. With a large drop like 25%, keep n in the denominator.

Answer from the NTA final answer key. Solution by PrepWiser.