Solutions and Colligative Properties - Result Question 35
####6. Liquids $A$ and $B$ form an ideal solution in the entire composition range. At $350 K$, the vapour pressures of pure $A$ and pure $B$ are $7 \times 10^{3} Pa$ and $12 \times 10^{3} Pa$, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of $A$ at this temperature is
(2019 Main, 10 Jan I)
(a) $x _A=0.76 ; x _B=0.24$
(b) $x _A=0.28 ; x _B=0.72$
(c) $x _A=0.4 ; x _B=0.6$
(d) $x _A=0.37 ; x _B=0.63$
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Solution:
- For ideal solution,
$$ \begin{aligned} p & =x _A^{\prime} p _A^{\circ}+x _B^{\prime} p _B^{\circ} \ \because \quad x _A^{\prime} & =0.4, x _B^{\prime}=0.6 \ p _A^{\circ} & =7 \times 10^{3} Pa, p _B^{\circ}=12 \times 10^{3} Pa \end{aligned} $$
On substituting the given values in Eq. (i), we get
$$ \begin{aligned} p & =0.4 \times 7 \times 10^{3}+0.6 \times 12 \times 10^{3} \ & =10 \times 10^{3} Pa=1 \times 10^{4} Pa \end{aligned} $$
In vapour phase,
$$ \begin{aligned} x _A & =\frac{p _A}{p}=\frac{x _A^{\prime} p _A^{\circ}}{p}=\frac{0.4 \times 7 \times 10^{3}}{1 \times 10^{4}}=0.28 \ \therefore \quad x _B & =1-0.28=0.72 \quad\left[\because x _A+x _B=1\right] \end{aligned} $$
