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At 300 K, a solution containing 1 mole of n-hexane and 3 moles of n-heptane has a vapour pressure of 550 mm Hg. At the same temperature, when one additional mole of n-heptane is added to the solution, the vapour pressure increases by 10 mm Hg. Assuming ideal behaviour, what is the vapour pressure (in mm Hg) of pure n-heptane?

1. 100 mm Hg
2. 110 mm Hg
3. 120 mm Hg
4. 130 mm Hg

Correct answer: 120 mm Hg

Solution

Using Raoult's law with mole fractions: first condition gives a/4 + 3b/4 = 550. Second condition (4 mol heptane, 1 mol hexane) gives a/5 + 4b/5 = 560. Multiplying the first equation by 4/5 and the second by 1: subtracting gives b = 120 mm Hg.

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