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^(@)`

Options

A. `97.5`
B. `90.75`
C. `105.5`
D. `85.75`

Answer

A

Explanation

`DeltaT_(f)` =Freezing point of water- Freezing point of solution `=0^(@)-(-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`
`alpha=0.975` ltbr Van't Hoff factor(i) = 2.95
Degree of dissociation = 0.975
Percentage degree of dissociation = 97.5