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The vapour pressures of benzene and toluene at 20^(@)C are 75mmHg and 22mmHg respectively. 23.4g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase if the vapours are in equilibrium with the liquid mixture at the same temperature. |
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Answer» Solution :No.of MOLES of BENZENE in liquid `=(23.4)/(78)=0.3` `(C_(6)H_(6)=78)` No. of moles of toluene in liquid `=(64.4)/(92)=0.7`. `(C_(6)H_(5)CH_(3)=92)` `:.` mole fraction of benzene `=(0.3)/(0.3+0.7)=0.3` `:.` mole fraction of toluene `=(0.7)/(0.3+0.7)=0.7` Partial pressure of benzene `=75xx0.3=22.5` Partial pressure of toluene `=22xx0.7=15.4`..........(Eqn.2) TOTAL pressure `=22.5+15.4=37.9mm` Mole fraction of benzene in vapour phase `=("partial pressure")/("total pressure")`..........(Eqn. 4) `=(22.5)/(37.9)=0.59` |
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