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A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given `K_(f)=1.86^(@)C" kg mol "^(-1) "for water "`
A. `-0.45^(@)C`
B. `-0.90^(@)C`
C. `-0.31^(@)C`
D. `-0.53^(@)C`

1 Answer

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Best answer
Correct Answer - a
HX ionises in solution
`HXhArrH^(+)+X^(-)`
`alpha=(i-1)/(n-1)or 0.2=(i-1)/(2-1)`
`DeltaT_(f)=ixxK_(f)xxm`
`=1.2 xx(1.86^(@)Cmol^(-1))xx(0.2 m)`
=`0.45^(@)C`
Freezing point of solution
`=0^(@)C-45^(@)C=-45^(@)C`.

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