The vapour pressure of benzene at a certain t | vedclass

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(C) Given,vapour pressure of pure benzene,$p^{\circ} = 640 \ mm \ Hg$. Vapour pressure of the solution,$p = 600 \ mm \ Hg$. Weight of solute,$w = 2.17

Vedclass · Accepted Answer

$69.60$

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(C) Given,vapour pressure of pure benzene,$p^{\circ} = 640 \ mm \ Hg$. Vapour pressure of the solution,$p = 600 \ mm \ Hg$. Weight of solute,$w = 2.175 \ g$. Weight of solvent (benzene),$W = 39.08 \ g$. Molecular weight of benzene,$M = 78 \ g/mol$. Let the molecular weight of the solute be $m$. According to Raoult's law for non-volatile solutes: $\frac{p^{\circ} - p}{p^{\circ}} = \frac{w 	imes M}{m 	imes W}$ Substituting the values: $\frac{640 - 600}{640} = \frac{2.175 	imes 78}{m 	imes 39.08}$ $\frac{40}{640} = \frac{169.65}{m 	imes 39.08}$ $\frac{1}{16} = \frac{169.65}{m 	imes 39.08}$ $m = \frac{16 	imes 169.65}{39.08} \approx 69.46 \approx 69.60 \ g/mol$.

The vapour pressure of benzene at a certain temperature is $640 \ mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \ g$ is added to $39.08 \ g$ of benzene. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?

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