Solutions and Colligative Properties 1 Question 4
5. The vapour pressures of pure liquids $A$ and $B$ are 400 and 600 $\mathrm{mmHg}$, respectively at $298 \mathrm{~K}$. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components $A$ and $B$ in vapour phase, respectively are
(2019 Main, 8 April I)
(a) $450 \mathrm{mmHg}, 0.4,0.6$
(b) $500 \mathrm{mmHg}, 0.5,0.5$
(c) $450 \mathrm{mmHg}, 0.5,0.5$
(d) $500 \mathrm{mmHg}, 0.4,0.6$
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Solution:
- (d) According to Dalton’s law of partial pressure
$$ \begin{aligned} p_{\text {total }} & =p_{A}+p_{B} \ & =p_{A}^{\circ} \chi_{A}+p_{B}^{\circ} \chi_{B} \end{aligned} $$
Given, $p_{A}^{\circ}=400 \mathrm{~mm} \mathrm{Hg}, p_{\mathrm{B}}^{\circ}=600 \mathrm{~mm} \mathrm{Hg}$
$$ \begin{aligned} & \chi_{B}=0.5, \chi_{A}+\chi_{B}=1 \ \therefore & \chi_{A}=0.5 \end{aligned} $$
On substituting the given values in Eq. (i). We get,
$p_{\text {total }}=400 \times 0.5+600 \times 0.5=500 \mathrm{~mm} \mathrm{Hg}$
Mole fraction of $A$ in vapour phase,
$$ Y_{A}=\frac{p_{A}}{p_{\text {total }}}=\frac{p_{A}^{\circ} \chi_{A}}{p_{\text {total }}}=\frac{0.5 \times 400}{500}=0.4 $$
Mole of $B$ in vapour phase,
$$ \begin{gathered} Y_{A}+Y_{B}=1 \ Y_{B}=1-0.4=0.6 \end{gathered} $$
