Solutions and Colligative Properties - Result Question 5-1
5. The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600$ $mm $ $Hg$ , 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 Aprill)
(a) $450 $ $mm $ $Hg, 0.4,0.6$
(b) $500 $ $mm $ $Hg, 0.5,0.5$
(c) $450 $ $mm$ $Hg, 0.5,0.5$
(d) $500$ $ mm$ $Hg, 0.4,0.6$
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Answer:
Correct Answer: 5. (d)
Solution:
(d) According to Dalton’s law of partial pressure
$\begin{aligned} p_{\text {total }} & =p_A+p_B \\ & =p_A^{\circ} \chi_A+p_B^{\circ} \chi_B \quad….(i) \end{aligned}$
Given, $p_A^{\circ}=400 \mathrm{mm} $ $\mathrm{Hg}, p_{\mathrm{B}}^{\circ}=600 \mathrm{mm} $ $\mathrm{Hg}$
$\begin{aligned} & \chi_B=0.5, \chi_A+\chi_B=1 \\ \quad & \chi_A=0.5 \end{aligned}$
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,
$\begin{gathered} Y_A+Y_B=1 \\ Y_B=1-0.4=0.6 \end{gathered}$
BETA
