Verified Solution • 2025–26 Session moderate

QJEEM20Ch4bba: JEE Main 2020 Chemistry - Class 12 JEEM20Ch4bba Solution: At 300 K, the vapour pressure of a solution containing 1 mole…

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 ______?

Step-by-Step Verified Solution

Let X1 and P P1oP_1^o are the mole fraction and vapour pressure of n-hexane in solution and X2 and P2oP_2^o are the mole fraction and vapour pressure of n-heptane in solution then

550 = P1oP_1^o ×\times 14{1 \over 4} + P2oP_2^o ×\times 34{3 \over 4}

⇒\Rightarrow 2200 = P1oP_1^o + 3 P2oP_2^o ....(1)

On addition of 1 more mole of n-heptane

560 = P1oP_1^o ×\times 15{1 \over 5} + P2oP_2^o ×\times 45{4 \over 5}

⇒\Rightarrow 2800 = P1oP_1^o + 4 P2oP_2^o ....(2)

From (1) and (2),

P1oP_1^o = 400, P2oP_2^o = 600

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