In an experiment, 1 g of a non-volatile solu | vedclass

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(C) According to Raoult's Law for a non-volatile solute, the relative lowering of vapour pressure is given by:$\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) According to Raoult's Law for a non-volatile solute, the relative lowering of vapour pressure is given by:$\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$Given: $P^o = 195 \ mm \ Hg$, $P_s = 192.5 \ mm \ Hg$, $w_2 = 1 \ g$, $w_1 = 100 \ g$, $M_1 = 58 \ g/mol$.Substituting the values:$\frac{195 - 192.5}{195} = \frac{1 	imes 58}{M_2 	imes 100}$$\frac{2.5}{195} = \frac{58}{100 	imes M_2}$$M_2 = \frac{58 	imes 195}{2.5 	imes 100} = \frac{11310}{250} = 45.24 \ g/mol$.The closest option is $45 \ g/mol$.

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