Two liquids $\dpi{100} X and \: Y$ 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 $\dpi{100} X and \: Y$ 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

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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