Question

At ${80}^{o}C$, the vapour pressure of pure liquid $A$' is $520mmHg$ and that of pure liquid $B$' is $1000mmHg$. If a mixture solution of $A$' and $B$' boils at ${80}^{o}C$ and 1 atm pressure, the amount of $A$' in the mixture is:

[$1atm=760mmHg$]

• A. $48mol\%$
• B. $50mol\%$
• C. $52mol\%$
• D. $34mol\%$

Answer 1

The correct option is B $50mol\%$

In a mixture of liquids of A and B,

$p_A=x_A\times {p_A}^0$

$p_B=x_B\times {p_B}^0$

Here, $p_A$ and $p_B$ are the partial vapour pressures of the components A and B.

The total vapour pressure of the mixture is equal to the sum of the individual partial pressures.

$p=p_A+p_{B}p=x_A.{p_A}^0+x_B.{p_B}^0$

Given,

${p_A}^0=520$ mm of Hg

${p_B}^0=1000$ mm of Hg

Resultant pressure is $p=760$ mm of Hg

Let mole fraction of A be $x_A$ and mole fraction of B will be $1-x_A$.

Then,

$760=520\times x_A+1000(1-x_A)$

$\Rightarrow x_A=1/2=0.5$

Hence, the solution contains $50\%$ of liquid A and $50\%$ of liquid B.

Hence, the correct option is $B$