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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 p0solvent−psolventp0solvent=i×χsoluteχsolute=moles of solutemoles of solute+moles of solventGiven that nNaCl=1 and nH2O=3 and p0solvent−psolventp0solvent=0.4Thus 0.4=i×11+3i=4×0.4=1.6Let the degree of dissociation of NaCl be α=fraction of molecules of NaCl dissociated=number of moles of NaCl dissociated/initial moles of NaCl presentFor 1mol of NaCl:NaCl→Na++Cl−1−ααα total number of moles after dissociation =1−α+α+α=1+α since i=effective number of molecules/initial number of moleculesi=(1+α)/1as above calculated, i=1.6thus1.6=1+α1α=1.6−1=0.6degree of dissociation =0.6and therefore % dissociation = 0.6×100 = 60 %

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