JEE MAIN - Chemistry (2020 - 4th September Morning Slot - No. 8)
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
600
Explanation
Let X1 and P$$P_1^o$$
are the mole fraction and vapour
pressure of n-hexane in solution and X2 and $$P_2^o$$
are the mole fraction and vapour pressure of
n-heptane in solution then
550 = $$P_1^o$$$$ \times $$$${1 \over 4}$$ + $$P_2^o$$$$ \times $$$${3 \over 4}$$
$$ \Rightarrow $$ 2200 = $$P_1^o$$ + 3$$P_2^o$$ ....(1)
On addition of 1 more mole of n-heptane
560 = $$P_1^o$$$$ \times $$$${1 \over 5}$$ + $$P_2^o$$$$ \times $$$${4 \over 5}$$
$$ \Rightarrow $$ 2800 = $$P_1^o$$ + 4$$P_2^o$$ ....(2)
From (1) and (2),
$$P_1^o$$ = 400, $$P_2^o$$ = 600
550 = $$P_1^o$$$$ \times $$$${1 \over 4}$$ + $$P_2^o$$$$ \times $$$${3 \over 4}$$
$$ \Rightarrow $$ 2200 = $$P_1^o$$ + 3$$P_2^o$$ ....(1)
On addition of 1 more mole of n-heptane
560 = $$P_1^o$$$$ \times $$$${1 \over 5}$$ + $$P_2^o$$$$ \times $$$${4 \over 5}$$
$$ \Rightarrow $$ 2800 = $$P_1^o$$ + 4$$P_2^o$$ ....(2)
From (1) and (2),
$$P_1^o$$ = 400, $$P_2^o$$ = 600
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