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 ________?

A
600
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B
600.0
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C
600.00
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Solution

## If X1 and P∘1 are the mole fraction and vapour pressure of n -hexane in solution and X2 and P∘2 are the mole fraction and vapour pressure of n- heptane in solution then 550=X1P∘1+X2P∘2 X1=14; X2=34 =P∘14+3P∘24⇒P∘1+3P∘2=2200 ...(1) On addition of 1 more mole of n - heptane X′1=15; X′2=45 560=X′1P∘1+X′2P∘2 =P∘15+4P∘25⇒P∘1+4P∘2=2800 ...(2) From (1) and (2), we get, P∘2=600 mm Hg

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