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Liquids in Liquids

For a binary ...

Question

For a binary liquid solution obeying Raoult's law at $300 K$ ,
$X_{A} = 0.75$ , Given: $P_{A}^{o} = 400 m m H g; P_{B}^{o} = 600 m m H g$
Then select correct statement

A

Vapour pressure of solution will be

450 m m

H g

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B

In distillate mol fration of

B

will be

2 / 3

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C

If distillate is removed & condensed then it is redistilled, then mole fraction of

A

in

I I

distillate is

4 / 7

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D

Normal boiling point of liquid solution should be more than

300 K

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Solution

The correct options are
A Vapour pressure of solution will be $450 m m$ $H g$
C If distillate is removed & condensed then it is redistilled, then mole fraction of $A$ in $I I$ distillate is $4 / 7$
D Normal boiling point of liquid solution should be more than $300 K$
$P_{S} = P_{A}^{o} . X_{A} + P_{B}^{o} . X_{B}$
$P_{S} = 450 m m H g .$

$(A)$ is correct
$P_{T} Y_{B} = P_{B}^{o} X_{B}$
$Y_{B} = 1 / 3$

$(B)$ is incorrect
If distillate is removed and condensed then again redistilled,
$Y_{A} \to X_{A}^{'} = 2 / 3, Y_{B} \to X_{B}^{'} = 1 / 3$
$P^{'} = \frac{1400}{3} m m H g$
So,
$P_{S}^{'} . Y_{A}^{'} = P_{A}^{o} X_{A}^{'}$
$^{'} Y_{A}^{'} = 4 / 7$

$(C)$ is correct
Since at $300 K$ , $P_{S} = 450 m m H g < P_{a t m} = 760 m m H g$
So for boiling point $T > 300 K$

$(D)$ is correct

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