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 ______?
Answer (integer)
600
Solution
Let X<sub>1</sub> and P$P_1^o$
are the mole fraction and vapour
pressure of n-hexane in solution and X<sub>2</sub> and $P_2^o$
are the mole fraction and vapour pressure of
n-heptane in solution then
<br><br>550 = $P_1^o$$\times$${1 \over 4}$ + $P_2^o$$\times$${3 \over 4}$
<br><br>$\Rightarrow$ 2200 = $P_1^o$ + 3$P_2^o$ ....(1)
<br><br>On addition of 1 more mole of n-heptane
<br><br>560 = $P_1^o$$\times$${1 \over 5}$ + $P_2^o$$\times$${4 \over 5}$
<br><br>$\Rightarrow$ 2800 = $P_1^o$ + 4$P_2^o$ ....(2)
<br><br>From (1) and (2),
<br><br>$P_1^o$ = 400, $P_2^o$ = 600
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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