The vapour pressure of pure benzene at a cert | vedclass

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(C) Given: $p^o = 0.850 \ bar$, $p = 0.845 \ bar$, $W_{benzene} = 39.0 \ g$, $M_{benzene} = 78 \ g/mol$, $w_{solid} = 0.5 \ g$.Using Raoult's law for

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$170$

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(C) Given: $p^o = 0.850 \ bar$, $p = 0.845 \ bar$, $W_{benzene} = 39.0 \ g$, $M_{benzene} = 78 \ g/mol$, $w_{solid} = 0.5 \ g$.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).Calculate moles of benzene: $n_{benzene} = \frac{39.0 \ g}{78 \ g/mol} = 0.5 \ mol$.Substitute values into the equation: $\frac{0.850 - 0.845}{0.850} = \frac{0.5 / M_{solid}}{0.5}$.Simplify: $\frac{0.005}{0.850} = \frac{1}{M_{solid}}$.Solve for $M_{solid}$: $M_{solid} = \frac{0.850}{0.005} = 170 \ g/mol$.

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