`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

Observed mol. Mass is obtained from the given data, i.e.,
`w_(2)="0.5 g, "w_(1)="100 g, "M_(1)="18 g mol"^(-1)" for "H_(2)O, DeltaT_(f)=0-(-0.24)=0.24^(@)C`
`M_(2)=(1000K_(f)w_(2))/(DeltaT_(f)xxw_(1))=("1000 g kg"^(-1)xx"1.86 K kg mol"^(-1)xx"0.5 g")/("0.24 K"xx"100 g")="38.75 g mol"^(-1)`
Calculated (theoretical) mol mass of KCl `=39+35.5="74.5 g mol"^(-1)`
`therefore" i"=("Calculated mol mass")/("Observed mol mass")=(74.5)/(38.75)=1.92`
Now, KCl dissociated as
`{:(,KCl,hArr,K^(+),+,Cl^(-)),("Initial moles","1 mole",,0,,0),("Moles after disso.",1-alpha,,alpha,,alpha):}`
`"Total no. of moles after dissociation "=1-alpha+alpha+alpha+=1+alpha" "therefore" "i=(1+alpha)/(1) or alpha=i-1=1.92 -1=0.92`
Percentage ionization `=0.92xx100=92%`.