Calculate van't Hoff-factor for 0.2 m aqueous solution of KCI which freezes at - 0.680°C.
`(K_(f)=1.86 K kg mol^(-1))`

Options

A. 3.72
B. 1.83
C. 6.8
D. 1.86

Answer

B

Explanation

Given, Molality = 0.2 m
`K_(1)=1.86 K kg mol^(-)`
`DeltaT_(f)=0.680^(@)C`=0.680 K
As, we know that
`DeltaT_(f)=ixxK_(f)xxm`
`0.680=ixx1.86xx0.2`
`(0.680)/(1.86xx0.2)=i`
1.83=i