A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg . The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

Solution

Chemistry ma’am · 4 steps

Raoult's law gives p_B° directly, and lower pure vapour pressure means less volatile.

1. Mole fractions. There are 1 + 3 = 4 moles in total, so x_A = 1/4 = 0.25 and x_B = 3/4 = 0.75.

   pₜₒₜₐₗ = x_A p_A° + x_B p_B°

2. Put in numbers. 500 = 0.25 × 200 + 0.75 p_B°, so 500 = 50 + 0.75 p_B°. This gives p_B° = 450/0.75 = 600 mm Hg.
3. Least volatile. A liquid that evaporates less easily has a lower vapour pressure. Pure A has 200 mm Hg and pure B has 600 mm Hg, so A is the least volatile.
4. So. p_B° = 600 mm Hg and the least volatile component is A, which is option (c).

The idea to keep. Each liquid contributes its mole fraction times its pure vapour pressure. Higher pure vapour pressure means more volatile, lower means less volatile.

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