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The vapour pressure of pure benzene is 6...

The vapour pressure of pure benzene is `639.7 mm Hg` and the vapour pressure of solution of a solute in benzene at the temperature is `631.9 mm Hg`. Calculate the molality of the solution.

Text Solution

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The correct Answer is:
Molality=`0.156`
Given,
Vapour pressure of pure benzene `(P^(@))=639.7 mm` of `Hg`
Vapour pressure of solution `(P)=631.9 mm` of `Hg`
According to Raoult's law
`(P^(@)-P)/(P^(@))=(W_(2) xx Mw_(1))/(Mw_(2) xx W_(1))`
where `W_(2)` and `W_(1)` = Weight of solute and solvent, respectively.
`Mw_(2)`,`Mw_(1)`=Molecular weight of solute and solvent, respectively.
`:. "Molality" =W_(2)/(Mw_(2) xx W_(1)) xx 1000`
=`(P^(@))-P/(P^(@))xx1/(Mw) xx 100`
=`(639.7-631.9)/639.7 xx 1/78 x1000`
=`(7.8 xx 1000)/(639.7 xx 78)=0.156 "mol" kg^(-1)`
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