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Question

# The vapour pressure of two pure liquids A and B, that forms an ideal solution, are 100 and 900 torr respectively at temperature T. This liquid solution of A and B is composed of 1 mole of A and 1 mole of B. What will be the pressure, when 1 mole of this mixture has been vaporized?

A
800 torr
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B
500 torr
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C
300 torr
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D
None of these
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Solution

## The correct option is C 300 torrAfter 1 mole of liquid is vapourised lets say mole fraction as well as the number of moles of A in the liquid is xA so for B will (1−XA) respectively In the vapour phase, the mole fraction as well as the number of moles of A and B will be (1−XA) and xA respectively From the ideal gas equation P=nRTV hence PA=P0AxA (1−xA)RTV=100xA (i) similarly for B xARTV=900(1−xA) (ii) from equation (i) and (ii) we will get xA=3/4 The total vapour pressure for this concentration will be PT=100×3/4+900×1/4 =300 torr

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