Question

At $80^{\circ} C$, the vapour pressure of pure liquid $A$ is $520 \,mm$ of $Hg$ and that of pure liquid $B$ is $1000 \,mm$ of $Hg$. If a mixture solution of $A$ and $B$ boils at $80^{\circ} C$ and $1$ atm pressure, the amount of $A$ in the mixture is $(1 atm =760 \,mm$ of $Hg )$

• A. 50 mol percent
• B. 52 mol percent
• C. 34 mol percent
• D. 48 mol percent

Answer 1

Answer: (A)

We have, $P_{A}^{\circ}=520 \,mm \,Hg$ and $P_{B}^{\circ}=1000 \,mm\, Hg$

Let mole fraction of $A$ in solution $=x_{A}$

and mole fraction of $B$ in solution $=x_{B}$

Then, at 1 atm pressure i.e., at $760\, mm \,Hg$,

$P_{A}^{\circ} x_{A}+P_{B}^{\circ} x_{B}=P_{A}^{\circ} x_{A}+P_{B}^{\circ}\left(1-x_{A}\right)=760\, mm \,Hg$

$\Rightarrow 520 \,x_{A}+1000-1000 x_{A}=760\, mm \,Hg$

$\Rightarrow x_{A}=\frac{1}{2}$ or $50 \,mol$ percent