    Question

# Two liquids A and B form ideal solutions. At 300K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550mmHg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10mmHg. Determine the vapour pressure of A and B in the pure states (in mmHg).

A
400,600
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B
500,500
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C
600,400
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D
None of these
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Solution

## The correct option is B 400,600According to Raoult's law, Ptotal=P0A×xA+P0B×xB1 mole of A and 3 mole of B corresponds to the mole fractions 0.25 and 0.75 respectively.Substitute values in the above expression.550=P0A×0.25+P0B×0.75 ......(1)When one mole of B is added, the mole fractions become 0.2 and 0.8 respectively.550+10=P0A×0.20+P0B×0.8 ......(2)Equations (1) and (2) are solved to obtain then vapour pressures of A and B in the pure state. P0A=400mmHgP0B=600mmHg  Suggest Corrections  0      Similar questions  Related Videos   Liquids in Liquids and Raoult's Law
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