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In a 0.2 molal aqueous solution of weak ...

In a `0.2` molal aqueous solution of weak acid `HX` (the degree of dissociation `0.3`) the freezing point is (given `K_(f) = 1.85 K molality^(-1)`):

A

`-0.26^(@)C`

B

`+48^(@)C`

C

`-0.48^(@)C`

D

`-0.36^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta T = K_(f) xx "molality" xx (1+alpha)`
`= 1.86 xx 0.2(1+0.3) = 0.48`
`:. F.pt. = 0 - 0.48 = 0.48^(@)C`
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