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Question

The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of Hg?

A
0.2
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B
0.4
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C
0.6
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D
0.8
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Solution

The correct option is C 0.6
The equation of relative lowering of vapour pressure which is given as:
p01p1p01=x2
where p01= vapour pressure of pure solvent
p1= vapour pressure of solution
p01p1= decrease in vapour pressure
x2= mole fraction of solute
So, when decrease in vapour pressure is 10 mm Hg, then using above equation, we get
10 mm Hgp01=0.2p01=10 mm Hg0.2=50 mm Hg
i.e. vapour pressure of pure solvent = 50 mm Hg
Now, when decrease in vapour pressure is 20 mm Hg, then using same equation
2050=x2x2=25=0.4
i.e. mole fraction of solute in solution = 0.4
So, the mole fraction of solvent = 1 - 0.4 = 0.6

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