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The freezing point of 0.2molal K(2)SO(4)...

The freezing point of `0.2`molal `K_(2)SO_(4)`is `-1.1^(@)C`. Calculate van't Hoff factor and percentage degree of dissociation of `K_(2)SO_(4).K_(f)`for water is `1.86^(@)`

A

`97.5`

B

`90.75`

C

`105.5`

D

`85.75`

Text Solution

Verified by Experts

The correct Answer is:
A

`DeltaT_(f)` = freezing point of water- Freezing point of solution `=0^(@)C-(-1.1^(@)C)=1.1^(@)`
we know that , `DeltaT_(f)=ixxKfxxm`
`1.1=ixx1.86xx0.2`
`i=(1.1)/(1.86xx0.2)=2.95`
but we know`i=1+(n-1)alpha`
`2.95=1+(3-1)alpha=1+2alpha`
Van't Hoff's factor (i)=`2.95`
Degree of dissociation=`0.975`
Percentage degree of dissociation=`97.5`
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