A non-volatile, non-electrolyte solid solute | vedclass

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(C) $1$. According to Raoult's Law for a dilute solution, the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_

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$3$

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(C) $1$. According to Raoult's Law for a dilute solution, the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$.$2$. Given: $P^0 = 760 	ext{ mm Hg}$, $P_s = 750 	ext{ mm Hg}$. So, $\frac{760 - 750}{760} = \frac{10}{760} = \frac{1}{76} = \frac{n_2}{n_1}$.$3$. Boiling point elevation is given by: $\Delta T_b = K_b 	imes m$, where $m$ is molality.$4$. $\Delta T_b = 320 - 319.5 = 0.5 	ext{ K}$.$5$. Molality $m = \frac{n_2}{w_1 	ext{ (in kg)}} = \frac{n_2}{0.04 	ext{ kg}}$.$6$. Substituting values: $0.5 = 0.3 	imes \frac{n_2}{0.04} \Rightarrow n_2 = \frac{0.5 	imes 0.04}{0.3} = \frac{0.02}{0.3} = \frac{1}{15} 	ext{ mol}$.$7$. Using the relation from step $2$: $\frac{n_2}{n_1} = \frac{1}{76} \Rightarrow n_1 = 76 	imes n_2 = 76 	imes \frac{1}{15} = 5.066 	ext{ mol}$.$8$. The nearest integer is $5$.

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