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Question

At 300 K, the vapour pressure of a solution containing 1 mole of nhexane and 3 moles of nheptane 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 ________?

A
600
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B
600.0
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C
600.00
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Solution

If X1 and P1 are the mole fraction and vapour pressure of n -hexane in solution and X2 and P2 are the mole fraction and vapour pressure of n- heptane in solution then

550=X1P1+X2P2

X1=14; X2=34

=P14+3P24P1+3P2=2200 ...(1)

On addition of 1 more mole of n - heptane

X1=15; X2=45

560=X1P1+X2P2

=P15+4P25P1+4P2=2800 ...(2)

From (1) and (2), we get,
P2=600 mm Hg

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