First slide

General discussion of solutions

Question

1 mol benzene (P $°$ _benzene = 42 mm) and 2 mol toluene (P $°$ _toulene = 36 mm) will have

Moderate

A

Total vapour pressure of 38 mm.
B

Mole fraction of vapour of benzene above liquid mixture is 7/19
C

Positive deviation from Raoult's law.
D

Negative deviation from Raoult's law.

Solution

Benzene and toluene $\Rightarrow$ Ideal solution.
$P_{Total} = P °_{A} X_{A} + P °_{B} X_{B} = 42 \times \frac{1}{3} + 36 \times \frac{2}{3} = 38 mm Hg {X^{V}}_{Benzene} = \frac{P_{Benzene}}{P_{Total}} = \frac{42 \times \frac{1}{3}}{38} = \frac{7}{19}$

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