When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?

Solution

Chemistry ma’am · 4 steps

Use the first case to find the pure solvent's vapour pressure, then reuse it in the second case.

1. Raoult's law. For a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute.

   (Δ p)/p° = xₛₒₗᵤₜₑ

2. Find pure vapour pressure. In the first case, 10/p° = 0.2, so p° = 50 mm Hg. This value belongs to the pure solvent, so it stays the same in the second case.
3. Second case. Now xₛₒₗᵤₜₑ = 20/50 = 0.4. The mole fractions add up to 1, so xₛₒₗᵥₑₙₜ = 1 - 0.4 = 0.6.
4. So. The mole fraction of the solvent is 0.6, which is option (d).

The idea to keep. Relative lowering of vapour pressure gives the solute's mole fraction, not the solvent's. Read carefully which one is asked, then use xₛₒₗᵥₑₙₜ = 1 - xₛₒₗᵤₜₑ.

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