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^@C-(-1.1^@C)=1.1^@C`
We know that,
`DeltaT_"f"=ixxK_"f"xxm`
`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`
Van’t Haff factor (i) = 2.95
Degree of dissociation = 0.975
Percentage degree of dissociation = 97.5