Solutions and Colligative Properties 1 Question 5
6. Liquids $A$ and $B$ form an ideal solution in the entire composition range. At $350 \mathrm{~K}$, the vapour pressures of pure $A$ and pure $B$ are $7 \times 10^{3} \mathrm{~Pa}$ and $12 \times 10^{3} \mathrm{~Pa}$, 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,
$$ \begin{aligned} p & =x_{A}^{\prime} p_{A}^{\circ}+x_{B}^{\prime} p_{B}^{\circ} \ \because \quad x_{A}^{\prime} & =0.4, x_{B}^{\prime}=0.6 \ p_{A}^{\circ} & =7 \times 10^{3} \mathrm{~Pa}, p_{B}^{\circ}=12 \times 10^{3} \mathrm{~Pa} \end{aligned} $$
On substituting the given values in Eq. (i), we get
$$ \begin{aligned} p & =0.4 \times 7 \times 10^{3}+0.6 \times 12 \times 10^{3} \ & =10 \times 10^{3} \mathrm{~Pa}=1 \times 10^{4} \mathrm{~Pa} \end{aligned} $$
In vapour phase,
$$ \begin{aligned} x_{A} & =\frac{p_{A}}{p}=\frac{x_{A}^{\prime} p_{A}^{\circ}}{p}=\frac{0.4 \times 7 \times 10^{3}}{1 \times 10^{4}}=0.28 \ \therefore \quad x_{B} & =1-0.28=0.72 \quad\left[\because x_{A}+x_{B}=1\right] \end{aligned} $$
