At 300 K, 0.06 kg of an organic solute is d | vedclass

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(C) 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} = \frac{W_2 	imes M_1}{M_2

Vedclass · Accepted Answer

$340$

Vedclass · Answer

(C) For a dilute solution, the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1} = \frac{W_2 	imes M_1}{M_2 	imes W_1}$ Given: $W_2 = 0.06 \ kg = 60 \ g$, $W_1 = 1 \ kg = 1000 \ g$, $P^0 = 3.78 \ kPa$, $P_s = 3.768 \ kPa$, $M_1 = 18 \ g \ mol^{-1}$ (for water). $\Delta P = P^0 - P_s = 3.78 - 3.768 = 0.012 \ kPa$. Rearranging the formula for $M_2$: $M_2 = \frac{W_2 	imes M_1 	imes P^0}{\Delta P 	imes W_1}$ $M_2 = \frac{60 	imes 18 	imes 3.78}{0.012 	imes 1000} = \frac{4082.4}{12} = 340.2 \ g \ mol^{-1}$ The closest integer value is $340 \ g \ mol^{-1}$.

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