Question

At $300\, K$, two pure liquids $A$ and $B$ have vapour pressures respectively $150 \,mm \,Hg$ and $100\, mm\, Hg$, In an equimolar liquid mixture of $A$ and $ B, $ the mole fraction of $B$ in the vapour mixture at this temperature is

• A. 0.6
• B. 0.5
• C. 0.8
• D. 0.4

Answer 1

Answer: (D)

Let the moles of $A=x$
$\therefore $ The moles of $B=x[\because$ mixture is equimolar.]
Mole fraction of $A=\frac{x}{x+x}=0.5=$ mole fraction of B.
Thus, $p_{T}=p_{A}^{o} X_{A}+p_{B}^{o} X_{B}$
$p_{T}=150 \times 0.5+100 \times 0.5$
$=125$
$\therefore $ Mole fraction of $B$ in vapour mixture
$=\frac{p_{B}^{o} X_{B}}{p_{T}}$
$=\frac{100 \times 0.5}{125}$