The vapour pressure of a saturated solution of sparingly soluble salt ( ${XCl}_{3}$ ) was 17.20 mm Hg at $27^{\circ} C$ . If the vapour pressure of pure $H_{2} O$ is 17.25 mm Hg at 300 K, what is the solubility of sparingly soluble salt ${XCl}_{3}$ in mole/litre?

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$8.08 \times 10^{- 2}$

$4.04 \times 10^{- 3}$

$2.02 \times 10^{- 2}$

$4.04 \times 10^{- 2}$

answer is A.

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Detailed Solution

The solubility of sparingly soluble salt ${XCl}_{3}$ in mole/litre is

$\begin{aligned} \frac{P_{A}^{\circ} - P_{δ}}{P_{s}} = \frac{n_{B}}{n_{A}} = \frac{4 S}{\frac{1000}{18}}; \frac{0.05}{17.20} = \frac{18 \times 4 \times S}{1000} \\ S = \frac{1000 \times 0.05}{17.20 \times 4 \times 18} = 4.04 \times 10^{- 2} \end{aligned}$