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A 0.1 molal solution of an acid is 4.5% ...

A 0.1 molal solution of an acid is `4.5%` ionized. Calculate freezing point. (molecular weight of the acid is 300 ). `K_(f)="1.86 K mol"^(-1)"kg".`

A

`-0.199^(@)C`

B

`2.00^(@)C`

C

`0^(@)C`

D

`-0.269^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
D
If acid is `4.5%` ionized then `alpha=0.45.`
`DeltaT_(f)="molality"xxK_(f)=0.1xx1.86=0.186`
`DeltaT_("exp")=DeltaT_(N)(1+alpha)=0.186(1+0.45)=0.296^(@)C`
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