The vapour pressure of pure benzene at a cert | vedclass

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(C) Given: $p_{benzene}^o = 0.850 \ bar$,$p_{solution} = 0.845 \ bar$,$W_{benzene} = 39.0 \ g$,$M_{benzene} = 78 \ g/mol$,$w_{solid} = 0.5 \ g$.Using

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$170$

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(C) Given: $p_{benzene}^o = 0.850 \ bar$,$p_{solution} = 0.845 \ bar$,$W_{benzene} = 39.0 \ g$,$M_{benzene} = 78 \ g/mol$,$w_{solid} = 0.5 \ g$.Using Raoult's law for non-volatile solutes: $\frac{p^o - p}{p^o} = \frac{n_{solid}}{n_{solid} + n_{benzene}} \approx \frac{n_{solid}}{n_{benzene}}$ (for dilute solutions).$n_{benzene} = \frac{39.0 \ g}{78 \ g/mol} = 0.5 \ mol$.$\frac{0.850 - 0.845}{0.850} = \frac{0.5 / M_{solid}}{0.5}$.$\frac{0.005}{0.850} = \frac{0.5}{M_{solid} 	imes 0.5} = \frac{1}{M_{solid}}$.$M_{solid} = \frac{0.850}{0.005} = 170 \ g/mol$.

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