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BITSAT 2020 Paper

Section: Chemistry

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Question : 64 of 150

Marks: +1, -0

The vapour pressure of two pure liquids

A

B

that form an ideal solution, are 400 and

800\;\text{mm}

of Hg respectively at a temperature

t^{\circ}\text{C}

. The mole fraction of

A

in a solution of

A

B

whose boiling point is

t^{\circ}\,\text{C}

will be

0.4
0.8
0.1
0.2

Solution:

V.P. of solution at

t^{\circ}\text{C}=760\;\text{mm}

[at b.p., V.P. of solution =atompheric pressure]

Thus

= P_A^\circ \cdot x_A + P_B^\circ x_B

or

P = P_A^0 \cdot x_A + P_B^\circ \cdot (1 - x_A)[\because x_A + x_B = 1]

or

760 = 400 X_A + 800(1 - X_A) \;\; [\because P = 760 \text{ mm of Hg}]

or

-800 + 760 = -400 x_A

or

-40 = -400 x_A

or

x_A = \frac{40}{400} = 0.1

Thus mole fraction in solution is 0.1

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