The vapour pressure of benzene at a certain t | 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^0 - P_s}{P^0} = \frac{n}{n + N}

Vedclass · Accepted Answer

$69.5$

Vedclass · Answer

(C) According to Raoult's law for a non-volatile solute,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{w 	imes M}{m 	imes W}$ Where: $P^0 = 640 \, mm \, Hg$ (Vapour pressure of pure benzene) $P_s = 600 \, mm \, Hg$ (Vapour pressure of solution) $w = 2.175 \, g$ (Mass of solute) $W = 39.08 \, g$ (Mass of benzene) $M = 78 \, g/mol$ (Molar mass of benzene,$C_6H_6$) $m$ = Molar mass of solute Substituting the values: $\frac{640 - 600}{640} = \frac{2.175 	imes 78}{m 	imes 39.08}$ $\frac{40}{640} = \frac{169.65}{m 	imes 39.08}$ $m = \frac{169.65 	imes 640}{39.08 	imes 40}$ $m = 69.45 \, g/mol$ Thus,the molecular weight of the solid substance is approximately $69.5 \, g/mol$.

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. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?

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