At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n-heptane in its pure state ______?

Solution

Chemistry ma’am · 3 steps

Two solutions, two unknowns: write Raoult's law for each mixture and solve the pair together.

1. First solution. With 1 mol hexane and 3 mol heptane, the mole fractions are 1/4 and 3/4. Raoult's law gives 550 = 1/4p_H° + 3/4p_(Hp)°, so p_H° + 3p_(Hp)° = 2200.
2. Second solution. Adding 1 mol heptane gives 1 mol hexane and 4 mol heptane, so the mole fractions become 1/5 and 4/5. The vapour pressure rises to 560 mm Hg, so p_H° + 4p_(Hp)° = 2800.

   (p_H° + 4p_(Hp)°) - (p_H° + 3p_(Hp)°) = 2800 - 2200 ⇒ p_(Hp)° = 600

3. So. Pure n-heptane has a vapour pressure of 600 mm Hg. As a check, pure n-hexane comes out at 2200 - 1800 = 400 mm Hg, and 1/4(400) + 3/4(600) = 550.

The common slip: 400. Solving for the pure vapour pressure of n-hexane instead of n-heptane gives 400, not 600.

The idea to keep. For an ideal mixture, each composition gives one Raoult's law equation. Two compositions means two equations, and that is enough to find both pure vapour pressures.

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