At T(K),the vapour pressure of pure benzene ( | vedclass

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(D) According to Raoult's law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$.Given: $P^o = 0.85 \ b

Vedclass · Accepted Answer

$0.784$

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(D) According to Raoult's law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$.Given: $P^o = 0.85 \ bar$,$P_s = 0.83 \ bar$,$W_2 = 2.0 \ g$,$W_1 = 39 \ g$,$M_1 = 78 \ g \ mol^{-1}$.$\frac{0.85 - 0.83}{0.85} = \frac{2.0 / M_2}{39 / 78}$.$\frac{0.02}{0.85} = \frac{2.0 / M_2}{0.5} \implies \frac{0.02}{0.85} = \frac{4}{M_2}$.$M_2 = \frac{4 	imes 0.85}{0.02} = 200 \ g \ mol^{-1}$.Now,calculate molality $(m)$: $m = \frac{W_2 	imes 1000}{M_2 	imes W_1(g)} = \frac{2.0 	imes 1000}{200 	imes 39} = \frac{10}{39} \approx 0.2564 \ mol \ kg^{-1}$.Elevation in boiling point: $\Delta T_b = K_b 	imes m = 2.6 	imes \frac{10}{39} = \frac{26}{39} = \frac{2}{3} \approx 0.667 \ K$. Wait,re-calculating: $\frac{0.02}{0.85} = \frac{n_2}{0.5} \implies n_2 = \frac{0.01}{0.85} \approx 0.01176 \ mol$.$M_2 = \frac{2.0}{0.01176} = 170 \ g \ mol^{-1}$.$m = \frac{0.01176}{0.039} = 0.3015 \ mol \ kg^{-1}$.$\Delta T_b = 2.6 	imes 0.3015 = 0.784 \ K$. Thus,the correct option is $D$.

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