On dissolving 0.5 g of a non-volatile non-io | vedclass

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(A) According to Raoult's law for relative lowering of vapor pressure:$\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_

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$1.01$

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(A) According to Raoult's law for relative lowering of vapor pressure:$\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$Given: $P^{\circ} = 650 \ mm \ Hg$,$P_s = 640 \ mm \ Hg$,$W_2 = 0.5 \ g$,$W_1 = 39 \ g$,$M_1 = 78 \ g \ mol^{-1}$,$K_f = 5.12 \ K \ kg \ mol^{-1}$.$\frac{650 - 640}{650} = \frac{0.5 / M_2}{39 / 78} = \frac{0.5 / M_2}{0.5} = \frac{1}{M_2}$$\frac{10}{650} = \frac{1}{M_2} \implies M_2 = 65 \ g \ mol^{-1}$.Now,calculate molality $(m)$:$m = \frac{W_2 	imes 1000}{M_2 	imes W_1} = \frac{0.5 	imes 1000}{65 	imes 39} \approx 0.197 \ mol \ kg^{-1}$.Depression in freezing point $\Delta T_f = K_f 	imes m = 5.12 	imes 0.197 \approx 1.01 \ K$.

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