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The vapour pressure of two liquids A an...

The vapour pressure of two liquids A and B in their pure states are in ratio of `1:2`. A binary solution of A and B contains A and B in the mole proportion of `1:2`. The mole fraction of A in the vapour phase of the solution will be

A

0.33

B

0.2

C

0.25

D

0.52

Text Solution

Verified by Experts

The correct Answer is:
B
Given that `(p_(A))/(p_(B))=1/2`
`:.p_(B)^(@)=2p_(A)^(@)`
Also, `(n_(A))/(n_(B))=1/2`
`:.x_(A)=(n_(A))/(n_(A)+n_(B))=(1)/(1+(n_(B))/(n_(A)))=(1)/((1+2))`
`=1/3=0.33`
`x_(B)=1-x_(A)=1-0.33=0.67`
Mole fraction of A in vapour phase of solution,
`y_(A)=(p_(A))/(p_("total"))=(p_(A)^(@)x_(A))/(p_(A)^(@)x_(A)+p_(B)^(@)x_(B))`
Substituting the values,
`y_(A)=(p_(A)^(@)(0.33))/(p_(A)^(@)(0.33)+2p_(A)^(@)(0.67))`
`=(0.33p_(A)^(@))/(p_(A)^(@)(0.33+1.34))=(0.33)/(1.67)=0.2`.
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