Question

The lowering of vapour pressure in a saturated aq. solution of salt $AB$ is found to be $0.108torr$. If vapour pressure of pure solvent at the same temperature is $300torr$, find the solubility product of salt $AB$:

• A. $4\times {10}^{-8}$
• B. $8\times {10}^{-6}$
• C. $4\times {10}^{-4}$
• D. $2\times {10}^{-5}$

Answer 1

Answer: (C)

$4\times {10}^{-4}$

The expression for the relative lowering in the vapour pressure for dilute solutions is $\frac{\Delta P}{P}=\frac{n}{N}$

Here. $\Delta P$ is the lowering in the vapor pressure, P is the vapor pressure of pure solvent, n is the number of moles of solute and N is the number of moles of solvent. Consider 1 L of the solution.

$\frac{0.108}{300}=\frac{n}{55.55}$

$n=\frac{0.108\times 55.55}{300}=0.02$.

Thus, the molar concentration (or solubility) is 0.02 M.

The solubility product is $S^2={0.02}^2=4\times {10}^{-4}$