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 10 mm Hg. Vapour pressure (in mm Hg) of X and \: Y in their pure states will be, respectively

  • Option 1)

    200 and 300

  • Option 2)

    300 and 400

  • Option 3)

    400 and 600

  • Option 4)

    500 and 600

 

Answers (1)

As we learnt in

Rault's Law -

The total vapour pressure of binary mixture of miscible liquids be having ideally is given by

P_{T}= P_{A}^{0}x_{A}+P_{B}^{0}x_{B}

Where x_{A}  and  x_{B} are mole fraction of A and B in liquid phase.

 

 

- wherein

P_{A}^{0}  and P_{B}^{0} are vapour pressures of pure liquids.

 

 p=p_{A}^{0}\chi_{A}+p_{B}^{0}\chi_{B}

Given that 1 mole x & 3 mole y 

\chi_{A}=\frac{1}{4},\ \; \chi_{B}=\frac{3}{4}

550=p_{A}^{0}\times \frac{1}{4}+p_{B}^{0}\times \frac{3}{4}                                        (1)

After adding 1 mole y the vapour pressure is 560 

560=p_{A}^{0}\times \frac{1}{5}+p_{B}^{0}\times \frac{4}{5}                                        (2)

By solving equation 1 & 2

p_{A}^{0}=400,\ \; p_{B}^{0}=600

Correct option is 3.

 


Option 1)

200 and 300

This is an incorrect option.

Option 2)

300 and 400

This is an incorrect option.

Option 3)

400 and 600

This is the correct option.

Option 4)

500 and 600

This is an incorrect option.

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