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The molar volume of liquid benzene (density =0.877g//mL) increases by a factor of 2750 as it vaporises at 20^(@)C and that of liquid toluene (density =0.867g//mL) increases by a factor of 7720 at 20^(@)C. A solution of benzene and toluene at 20^(@)C has a vapour pressure of 46 torr. Find the mole fraction of benzene in the vapour above the solution |
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Answer» <P> Solution :The volume of 1 mole of `C_(6)H_(6)` (liquid) `=(78)/(0.877)mL`The volume of 1 mole of `C_(6)H_(6)` (VAPOUR) `=(78)/(0.877)xx2750mL` `=244.58` LITRES for 1 mole of `C_(6)H_(6)` vapour at `20^(@)C` `p_(b)^(0)V=RT` `p_(b)^(0)=(0.0821xx293)/(244.58)`atm`=74.75mm` Similarly for 1 mole of `C_(6)H_(5)CH_(3)` vapour at `20^(@)C` `p_(t)^(0)=(0.0821xx293)/(((92)/(0.867)xx(7720)/(1000)))"atm"=22.32mm` Now, `(1)/(p)=(x_(b).)/(p_(b)^(0))+((1-x_(b).)/(p_(t)^(0)))` `(1)/(46)=(x_(b).)/(74.75)+(1-x_(b).)/(22.32)` `x_(b)^(.)=0.7336` |
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