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Question

If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol H2O. NaCl is ______ % ionised.

A
60%
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B
50%
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C
100%
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D
40%
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Solution

The correct option is C 60%
Relative decrease in vapour pressure p0solventpsolventp0solvent=i×χsolute
χsolute=moles of solutemoles of solute+moles of solvent
Given that nNaCl=1 and nH2O=3 and p0solventpsolventp0solvent=0.4
Thus 0.4=i×11+3
i=4×0.4=1.6
Let the degree of dissociation of NaCl be α=fraction of molecules of NaCl dissociated=number of moles of NaCl dissociated/initial moles of NaCl present
For 1mol of NaCl:
NaClNa++Cl
1ααα
total number of moles after dissociation =1α+α+α=1+α
since i=effective number of molecules/initial number of molecules
i=(1+α)/1
as above calculated, i=1.6
thus1.6=1+α1
α=1.61=0.6
degree of dissociation =0.6
and therefore % dissociation = 0.6×100
= 60 %

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