Solutions and Colligative Properties - Result Question 4

####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:

  1. (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,

$$ \begin{gathered} Y _A+Y _B=1 \ Y _B=1-0.4=0.6 \end{gathered} $$



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