An aqueous solution is prepared by dissolving | vedclass

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(A) Given: $n_{	ext{salt}} = 0.1 \ mol$,$W_{	ext{water}} = 1.8 \ kg = 1800 \ g$,$T = 35^{\circ} C$,$\alpha = 0.90$,$P_s = 59.724 \ mm \ Hg$,$P^{\cir

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given: $n_{	ext{salt}} = 0.1 \ mol$,$W_{	ext{water}} = 1.8 \ kg = 1800 \ g$,$T = 35^{\circ} C$,$\alpha = 0.90$,$P_s = 59.724 \ mm \ Hg$,$P^{\circ} = 60.000 \ mm \ Hg$.Moles of water $n_w = \frac{1800 \ g}{18 \ g/mol} = 100 \ mol$.Let $x$ be the number of ions per formula unit. The salt dissociates as $AB_x ightarrow A^{x+} + xB^-$. The van't Hoff factor $i = 1 + \alpha(x - 1) = 1 + 0.9(x - 1) = 0.1 + 0.9x$.Using Raoult's Law: $\frac{P^{\circ} - P_s}{P_s} = \frac{i \cdot n_{	ext{salt}}}{n_w}$.$\frac{60.000 - 59.724}{59.724} = \frac{(0.1 + 0.9x) \cdot 0.1}{100}$.$\frac{0.276}{59.724} = \frac{0.01 + 0.09x}{100}$.$0.004621 \approx \frac{0.01 + 0.09x}{100} \Rightarrow 0.4621 = 0.01 + 0.09x$.$0.4521 = 0.09x \Rightarrow x = \frac{0.4521}{0.09} \approx 5.02$.Thus,the number of ions per formula unit is $5$.

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