Solutions and Colligative Properties - Result Question 34
####5. The vapour pressures of pure liquids $A$ and $B$ are 400 and 600 $mmHg$, respectively at $298 K$. 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) $450 mmHg, 0.4,0.6$
(b) $500 mmHg, 0.5,0.5$
(c) $450 mmHg, 0.5,0.5$
(d) $500 mmHg, 0.4,0.6$
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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, $p _A^{\circ}=400 mm Hg, p _B^{\circ}=600 mm Hg$
$$ \begin{aligned} & \chi _B=0.5, \chi _A+\chi _B=1 \ \therefore & \chi _A=0.5 \end{aligned} $$
On substituting the given values in Eq. (i). We get,
$p _{\text {total }}=400 \times 0.5+600 \times 0.5=500 mm Hg$
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,
$$ \begin{gathered} Y _A+Y _B=1 \ Y _B=1-0.4=0.6 \end{gathered} $$
