Question

The vapour pressure of two pure liquids $A$ and $B$ that form an ideal solution, are $400$ and $800 \,mm$ of $Hg$ respectively at a temperature $t^{\circ} C$. The mole fraction of $A$ in a solution of $A$ and $B$ whose boiling point is $t^{\circ} C$ will be

• A. 0.4
• B. 0.8
• C. 0.1
• D. 0.2

Answer 1

Answer: (C)

V.P. of solution at $t^{\circ} C=760\, mm$
[at b.p., V.P. of solution = atmospheric pressure]
Thus $=P_{A}{ }^{\circ} \cdot x_{A}+P_{B}{ }^{\circ} x_{B}$
or $P=P_{A}{ }^{\circ} \cdot x_{A}+P_{B}{ }^{\circ} \cdot\left(1-x_{A}\right)$
$\left[\because x_{A}+x_{B}=1\right]$
or $760=400\, X_{A}+800\left(1-X_{A}\right)$
$[\because P=760 \,mm$ 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 $0.1$