JEE Main 2020 — Solutions Question with Solution
From: JEE Main 2020 (Online) 4th September Morning Slot
Question
At 300 K, the vapour pressure of a solution
containing 1 mole of n-hexane and 3 moles of
n-heptane is 550 mm of Hg. At the same
temperature, if one more mole of n-heptane is
added to this solution, the vapour pressure of
the solution increases by 10 mm of Hg. What is
the vapour pressure in mm Hg of n-heptane in
its pure state ______?
Enter your answer
Show full solutionCorrect answer: 600
Correct answer
600
Step-by-step explanation
Let X1 and P
are the mole fraction and vapour
pressure of n-hexane in solution and X2 and
are the mole fraction and vapour pressure of
n-heptane in solution then
550 = +
2200 = + 3 ....(1)
On addition of 1 more mole of n-heptane
560 = +
2800 = + 4 ....(2)
From (1) and (2),
= 400, = 600
550 = +
2200 = + 3 ....(1)
On addition of 1 more mole of n-heptane
560 = +
2800 = + 4 ....(2)
From (1) and (2),
= 400, = 600
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