Benzoic acid completely demerises in benzene. What will be the vapour pressure of solution containing 61 g of benzoic acid per 500 g benzene when the vapour pressure of pure benzene at the temperature of experiment is 66.6 torr?

Solution

C_{7} H_{5} COOH mol . mass = 12 \times 6 + 5 \times 1 + 12 + 2 \times 16 + 1 = 122 g {mol}^{- 1} . C_{6} H_{5} COOH \to \frac{1}{2} (C_{6} H_{5} COO)_{2}

No. of moles after dissociation

1 - α α / 2

Total number of moles after association

= 1 - α + \frac{α}{2} = 1 - \frac{α}{2}

Mole fraction of benzoic acid

(n_{A}) = \frac{61}{122} = 0.5

Mole fraction of benzoic

(n_{B}) = \frac{500}{78} = 6.4 .

\frac{{p^{o}}_{A} - p_{A}}{{p^{o}}_{A}} = \frac{n_{A}}{n_{A} + n_{B}} \frac{66.6 - p_{A}}{66.6} = \frac{0.5}{0.5 + 6.4} = \frac{0.5}{6.9} = 0.072

66.6 - p_{A} = 66.6 \times 0.72 = 4.79

66.9 - 4.79 = p_{A}

∴

p_{A} = 61.8 torr

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