$224 m L$ of $S O_{2} (g)$ at $298 K$ and $1 a t m$ is passed through $100 m L$ of $0.1 M N a O H$ solution. The non-volatile solute produced is dissolved in $36 g$ of water. The lowering of vapour pressure of solution (assuming the solution is dilute) $(P_{H_{2} O} = 24 m m H g)$ is $x \times 10^{- 2} m m H g,$ the value of $x$ is _____.

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Solution

The balanced equation is:
$S O_{2} + 2 N a O H \to N a_{2} S O_{3} + H_{2} O$
$Moles of N a O H = molarity \times volume (in L) = 0.1 \times 0.1 = 0.01 m o l$

$Moles of S O_{2} = \frac{P V}{R T} = \frac{1 \times 224}{1000 \times 0.0821 \times 298} \approx 0.1 m o l$

According to equation, $1 m o l S O_{2}$ reacts with $2 m o l N a O H .$ Hence, $0.1 m o l S O_{2}$ should reacts with $0.2 m o l N a O H .$ Thus, $N a O H$ is a limiting reagent.

$2 m o l N a O H \equiv 1 m o l N a_{2} S O_{3} 0.01 m o l N a O H \equiv 0.005 m o l N a_{2} S O_{3}$

$N a_{2} S O_{3} \to 2 N a^{+} + S {O^{2 -}}_{3}$
$∴ i = 3$

$Moles of H_{2} O (solvent) = \frac{36}{18} = 2 m o l$

According to RLVP,
$\frac{P_{A}^{\circ} - P_{A}}{P_{A}^{\circ}} = i X_{B}$

$\frac{P_{A}^{\circ} - P_{A}}{P_{A}^{\circ}} = i \frac{n_{B}}{n_{A} + n_{B}}$

$∵ n_{B} << n_{A}$
$∴ n_{A} + n_{B} \approx n_{A}$

$\frac{P_{A}^{\circ} - P_{A}}{P_{A}^{\circ}} = i \frac{n_{B}}{n_{A}}$

$\frac{24 - P_{A}}{24} = 3 \times \frac{0.005}{2}$
$\Rightarrow 24 - P_{A} = 0.18$
$Lowering in vapour pressure = 0.18 m m H g = 18 \times 10^{- 2} m m H g$

$∴ x = 18$

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