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Question

Two liquids X and Y form an ideal solution.At 300 K,vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg.At the same temperature, if 1 mol of Y is further added to this solution,vapour pressure of the solution increases by 10mm Hg.So, vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively,

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Solution

The correct option is **A**

400 and 600

Mole fraction of X = 11+3=14

Mole fraction of Y = 31+3=34

Total pressure = Partial pressure of X * Mole fraction of X + partial pressure of Y * mole fraction of Y

550=Px4+3Py4 ------------1

2200=P∘x+3P∘y^{ }

When one mole of Y is added then mole fraction of X becomes 11+4=15 and mole fraction of Y becomes 41+4=45 . Now the total pressure given as

560=Px5+4Py5 ------------2

2800=P∘x+4P∘y^{ }

Solving eqns 1 and 2 , we get P∘x_{ }= 400 mm Hg and P∘y = 600 mm Hg.

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