In a solution of 7.8 g benzene (C(6)H(6)...

In a solution of 7.8 g benzene `(C_(6)H_(6))` and 46.0 g toluene `(C_(6)H_(5)CH_(3))` the mole fraction of benzene is :

`1//6`

`1//5`

`1//2`

`1//3`

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The correct Answer is:

Mole fraction of `C_(6)H_(6)=(7.8//78)/(7.8/78+46/92)=1/6`

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At 25^@C the vapour prssure of benzene , C_(6)H_(6)(78g//mol e) ,is 93.2 torr and that of toluene , C_(7)H_(8)(92g//mol) ,is 28.2 torr .A solution of 1.0 mole fo C_(6)H_(6) and 1.0 mol of C_(7)H_(8) id prepared. Calculate the mole fraction of C_(6)H_(6) in the vapour above this solution (assume the solution is ideal.)

At 298 K, the vapour pressure of pure benzene , C_(6)H_(6) is 0.256 bar and the vapour pressure of pure toluene C_(6)H_(5)CH_(3) is 0.0925 bar . If the mole fraction of benzene in solution is 0.40 (i) what is the total vapour pressure of the solution ? (ii) Calculate the composition of the vapour in terms of mole fraction.

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The molar volume of liquid benzene (density 0.877 g mL^(-1)) increases by a factor of 2750 as it vapourises at 20^(@)C and that of liquid toluene(density 0.867 g mL) increases by a factor of 7720 at 20^(@)C . A solution of benzene and toluene at 20^(@)C has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution.

At 334K the vapour pressure of benzene (C_(6)H_(6)) is 0.526 atm and that of toluene (C_(7)H_(8)) is 0.188 atm. In a solution containing 0.500 mole of benzene and 0.500 mole of toluene , what is the vapour pressure of toluene above the solution at 334K ?

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