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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? |
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Answer» <P> 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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