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A 0.2 molal aqueous solution of a weak a...

A `0.2` molal aqueous solution of a weak acid `HX` is `20%` ionized. The freezing point of the solution is `(k_(f) = 1.86 K kg "mole"^(-1)` for water):

A

`-0.45^(@)C`

B

`-0.90^(@)C`

C

`-0.31^(@)C`

D

`-0.53^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
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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