Question

The vapour pressure of a saturated solution of sparingly soluble salt $\left(XC{l}_3\right)$ was $17.20mmHg$ at $27^{\circ }C$. If the vapour pressure of pure $H_{2}O$ is $17.25mmHg$ at the same temperature, what is the solubility of sparingly soluble salt $XC{l}_3$, in moles per ltr. (Assume solution to be dilute)

• A. $4.02\times 10^{-2}$
• B. $6\times 10^{-7}$
• C. $8\times 10^{-5}$
• D. $2\times 10^{-4}$

Answer 1

Answer: (A)

$\underset{(s)}{XC{l}_3}\rightleftharpoons X^{3+}+3C{l}^{-}$

Solubility $=s$

molality $=$ molarity

Apply, $\frac{P^0-P_s}{P^{\circ }}=\frac{{\text{ in }}_B}{i{n}_B+n_A}$

$\frac{17.25-17.20}{17.20}=\frac{i{n}_B}{n_A}$

$(\because$ solution is very dilute $)$

$\frac{0.05}{17.20}=\frac{4}{55.55}\times s$

$S=4.04\times 10^{-2}$

Since solution is dilute, $m=$ molarity

So, solubility $=4.03\times 10^{-2}$