Question

At $300 K$ the vapour pressures of two pure liquids. $A$ and $B$ are $100$ and $500$ mm $Hg$, respectively. If in a mixture of $A$ and $B$, the vapour pressure is $300 \,mm \,Hg$, the mole fractions of $A$ in the liquid an d in the vapour phase, respectively are

• A. 1/2 and 1/10
• B. 1/4 and 1/6
• C. 1/4 and 1/10
• D. 1/2 and 1/6

Answer 1

Answer: (D)

In liquid phase, according to Dalton’s law of partial pressure
$P_T =P_A +P_B$
$= x_A p^{\circ}_A$ + $x_Bp^{\circ}_B$
$= x_A p^{\circ}_A + (1-x_A)p^{\circ}_B$
$P_T = (p^{\circ}_A-p^{\circ}_B)x_A+p^{\circ}_B$
$300 = (-400)x_A + 500$
$x_{A}=\frac{1}{2} $
In vapour phase,
$x_{A}=\frac{p_{A}}{p_{T}}=\frac{p^{\circ}A x A}{p^{\circ}A x A+p^{\circ}B x B}$
$=\frac{100\times\frac{1}{2}}{100\times\frac{1}{2}+500\times\frac{1}{2}}$
$ =\frac{50}{300}=\frac{1}{6}$