Question

A saturated aqueous solution of sparingly soluble salt ${AB}_3$ has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at ${25}^{0}C$. The solubility product of ${AB}_3$ is:

• A. $1.087\times {10}^{-2}$
• B. $1.48\times {10}^{-4}$
• C. $5.35\times {10}^{-5}$
• D. none of these

Answer 1

The correct option is D none of these

moles fraction of salute $=\frac{△P}{P_0}$

$=\frac{0.08}{17.33}$

$A{B}_3\left(g\right)\rightleftharpoons A^{3+}\left(aq\right)+3B^{-}\left(aq\right)$

$\therefore$ Total = 48

$\therefore 48$ moles per litre water

$\therefore$ moles of $H_{2}O=\frac{1000}{18}$

$\therefore$ mole fraction $=\frac{45}{45+\frac{1000}{18}}$

$\approx \frac{45}{1000/18}$ $(\because 45<<\frac{1000}{18})$

$\therefore \frac{45\times 18}{1000}=\frac{0.08}{17.33}$

$\Rightarrow S=\frac{0.08\times 1000}{17.33\times 45\times 18}$

$\approx 0.57\times {10}^{-2}$

$=5.7\times {10}^{-3}$