Solutions and Colligative Properties - Result Question 15
16. Liquids $A$ and $B$ form ideal solution over the entire range of composition. At temperature $T$, equimolar binary solution of liquids $A$ and $B$ has vapour pressure $45$ torr. At the same temperature, a new solution of $A$ and $B$ having mole fractions $x _A$ and $x _B$, respectively, has vapour pressure of $22.5$ torr. The value of $x _A / x _B$ in the new solution is _______.
(Given that the vapour pressure of pure liquid $A$ is $20$ Torr at temperature $T$ )
(2018 Adv. Paper-1)
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Answer:
Correct Answer: 16. $(19)$
Solution:
Key Idea: Use the formula
$ p _{\text {Total }}=p _A^{\circ} \times \chi _A+p _B^{\circ} \times \chi _B $
and for equimolar solutions $\chi _A=\chi _B=\frac{1}{2}$
Given, $p _{\text {Total }}=45$ torr for equimolar solution
$ p _A^{\circ}=20 \text { torr } $
So, $\quad 45=p _A^{\circ} \times \frac{1}{2}+p _B^{\circ} \times \frac{1}{2}=\frac{1}{2}\left(p _A^{\circ}+p _B^{\circ}\right)$
or $p _A^{\circ}+p _B^{\circ}=90$ torr $\quad$…..(i)
But we know $p _A^{\circ}=20$ torr
$ \text { so, } \quad p _B^{\circ}=90-20=70 \text { torr } $ (From Eq. (i))
Now, for the new solution from the same formula
Given, $p_{\text {Total }}=22.5$ torr
So, $22.5=20 \chi_A+70\left(1-\chi_A\right)$ $\quad \left(\operatorname{As} \chi_A+\chi_B=1\right)$
or $22.5=70-50 \chi_A$
So, $ \chi_A=\frac{70-22.5}{50}=0.95 $
Thus $ \chi_B=1-0.95=0.05 \quad$ (as $\left.\chi_A+\chi_B=1\right)$
Hence, the ratio
$ \frac{\chi_A}{\chi_B}=\frac{0.95}{0.05}=19 $
BETA
