224 mL of $S O_{2 (g)}$ at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH 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 ) ( $P_{(H_{2} O)} = 24 m m o f H g$ ) is $x \times 10^{- 2}$ mm of Hg , the value of x is (Integer answer )

24.0

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Solution

Using $P V = n R T$
$n_{S O_{2}} = \frac{1 \times 0.224}{0.082 \times 298} = 0.0092 = 0.01 m o l e s$

$N a O H + S O_{2} \to N a H S O_{3}$

$n_{N a H S O_{3}} = 0.01$

$N a H S O_{3} \to N a^{+} + H S O_{3}^{-}$

vant't Hoff factor:
(i) = 2

$\frac{P_{H_{2} O}^{0} - P_{S}}{P_{H_{2} O}^{0}} = i \times \frac{n_{N a H S O_{3}}}{n_{H_{2} 0} + i n_{N a H S O_{3}}}$

$(a s n_{H S O_{3}} << n_{H_{2} O})$

$Lowering in vapour pressure = \frac{2 \times 0.01}{2 + 2 \times 2 \times 0.01} \times 24$

$Lowering in vapour pressure = 23.76 \times 10^{- 2} m m H g \approx 24 \times 10^{- 2} m m H g$

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