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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2018 · Paper 1 · Chemistry · Solutions

Liquids A and B form ideal solution over the entire range of...

Liquids A and B form ideal solution over the entire range of composition. At temperature TT, equimolar binary solution of liquids A and B has vapour pressure 45Torr45Torr . At the same temperature, a new solution of A and B having mole fractions xAandxB{x}_{A}and{x}_{B}, respectively, has vapour pressure of 22.5Torr22.5Torr. The value of xAxB\dfrac{{x}_{A}}{{x}_{B}}, in the new solution is ____.
(Given that the vapour pressure of pure liquid A is 20Torr20Torr at temperature TT)

Correct answer

0

Step-by-step solution

45 = {P}_{A}^{o} \times \dfrac{1}{2} + {P}_{B}^{o} \times \dfrac{1}{2}<br/><br />{P}_{A}^{o} + {P}_{B}^{o} = 90........\left(i\right)<br/>Given,<br />Given,{P}_{A}^{o} = 20Torr<br/><br />{P}_{B}^{o} = 70Torr<br/><br />⇒ 22 .5 torr = 20{x}_{A} + 70 \left(1 - {x}_{A}\right)
=70−50xA= 70 - 50{x}_{A}.
{x}_{A} = \left(\dfrac{70 - 22 .5}{50}\right)=0.95= 0 .95
xB=0.05{x}_{B} = 0 .05
So, XAXB=0.950.05=19\dfrac{{X}_{A}}{{X}_{B}} = \dfrac{0 .95}{0 .05} =19.
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