Raoult’s law design question:\nA non-volatile solute lowers a solvent’s vapor pressure by 10 mm Hg when its mole fraction is 0.2. To achieve a 20 mm Hg lowering under the same conditions, what should be the solvent mole fraction?

Introduction / Context:
Vapor-pressure lowering by a non-volatile solute is a classic colligative property described by Raoult’s law for ideal or near-ideal solutions. Engineers use this proportionality for estimating boiling point elevation, designing evaporators, and checking composition from pressure data.

Given Data / Assumptions:

• Initial vapor-pressure lowering ΔP_1 = 10 mm Hg at x_solute,1 = 0.2.
• Target vapor-pressure lowering ΔP_2 = 20 mm Hg.
• Same temperature and same solvent; non-volatile solute; ideal-solution assumption.

Concept / Approach:
For an ideal solution with a non-volatile solute, vapor-pressure lowering ΔP = x_solute * P^0, where P^0 is the pure-solvent saturation pressure at the temperature. Because P^0 is constant at fixed T, ΔP is directly proportional to the solute mole fraction. Thus, doubling ΔP requires doubling x_solute, and the solvent mole fraction is x_solvent = 1 − x_solute.

Step-by-Step Solution:
From ΔP ∝ x_solute, compute x_solute,2 = (ΔP_2 / ΔP_1) * x_solute,1 = (20/10)*0.2 = 0.4.Then x_solvent,2 = 1 − 0.4 = 0.6.Therefore, the solvent mole fraction should be 0.6.

Verification / Alternative check:
Using the more explicit Raoult relation: P_solvent = x_solvent * P^0 → ΔP = P^0 − x_solvent P^0 = (1 − x_solvent)P^0 = x_solute P^0. Doubling ΔP doubles x_solute, so the result is consistent.

Why Other Options Are Wrong:
0.2 and 0.1 correspond to smaller solvent fractions than required; 0.4 would not double ΔP; 0.8 would halve ΔP relative to the initial case.

Common Pitfalls:
Confusing solute and solvent fractions; forgetting that P^0 is constant at constant temperature.

Final Answer:
0.6