Question

The vapour pressure of a saturated solution of sparingly soluble salt $\left( XCl _{3}\right)$ was $17.20 \,mm \,Hg$ at $27^{\circ} C$. If the vapour pressure of pure $H _{2} O$ is $17.25 \,mm \,Hg$ at the same temperature, what is the solubility of sparingly soluble salt $XCl _{3}$, 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)

$\underset{(s)}{XCl _{3}} \rightleftharpoons X ^{3+}+3 Cl ^{-}$

Solubility $= s$

molality $=$ molarity

Apply, $\frac{ P ^{0}- P _{ s }}{ P ^{\circ}}=\frac{\text { in }_{ B }}{ in _{ B }+ n _{ A }}$

$\frac{17.25-17.20}{17.20}=\frac{ in _{ B }}{ n _{ A }}$

$(\because$ 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 $=4.03 \times 10^{-2}$