`0.5 g` `KCl` was dissolved in `100 g` water, and the solution, originally at `20^(@)C` froze at `-0.24^(@)C`. Calculate the percentage ionization of salt. `K_(f)` per `1000 g` of water =`1.86^(@)C`.

Explanation

Calculation of the observed molar mass of KCI
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
`W_(B)=0.5g , W_(A)=100g , K_(f)=1.86 K kg mol^(-1)`,
`DeltaT_(f)=0-(-0.24)=0.24^(@)C=0.24K`
`M_(B)=((1.86K kg mol^(-1))xx(0.5g))/((0.24 K)xx(0.1 Kg))=38.75 g mol^(-1)`
Calculation of Van't Hoff factor
`i=("Normal molar mass")/("Observed molar mass")=((74.5g mol^(-1)))/((38.75g mol^(-1)))=1.923`
Calculation of degree of ionization
Potassium (KCI) ionizes in solution as:
`KCIoverset("aq")toK^(+)(aq)CI^(-)(aq)`
Value of n=2, i=1.923,
Degree of ionization `(alpha)=((i-1))/((n-1))=((1.923-1))/((2-1))=0.923`
Percentage ionization=0.923xx 100=92.3%