1.

The vapour pressure of pure benzene(C_6H_6)at a given temperature is 640 mm Hg . 2.2 g of non -volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600mm Hg . Calcultaed the molar mass of the solute?

Answer»

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Solution :`P _(C _(6)H_(6))^(0) = 640mm Hg, W _(2) =2.2 g` (non volatile solute),` W _(1) = 40 g` (benzene)
`P _("solution") = 600 mm Hg, M _(2)=` ?
`(P ^(0) - P )/(P ^(0))= x _(2)`
` (640-600)/(640) = (n _(2) )/(n_(1) + n _(2)) [ because n _(1) GT gt n _(2) , n _(1) + n _(2) ~~n _(1) ]`
`(40)/(640) = (n _(2))/( n _(1))`
` 0.0625 = (W_(2) xx M _(1))/( M _(2) xx W_(1))`
`M _(2) = ( 2.2 xx 78)/( 0.0625 xx 40) = 68. 64 g mol ^(-1)`


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