A `0.5%` aqueous solution of `KCl` was found to freeze at `-0.24^(@)C`. Calculate the Van,t Hoff factor and degree of dissociation of the solute at this concentration. (`K_(f)` for water =`1.86 K kg mol`^(-1)`)

Explanation

Observed mol. Mass of KCl `= (100xxK_(f)xxw)/(DeltaTxxW)`
Given, `K_(f)=1.86"kg mol"^(-1),w=0.5g,W=100g,DeltaT=0.24`
So, Observed mol. Mass of KCl `= (1000xx1.86xx0.5)/(0.24xx100)=38.75`
Normal. mol. mass of `KCl=39+35.5=74.5`
van't Hoff factor `= ("Normal mol. mass")/("Observed mol. mass")`
`=(74.5)/(38.75)=1.92`
`underset((1-alpha))(KCl)hArrunderset(alpha)(K^(+))+underset(alpha)(Cl^(-))` (`alpha` is the degree of ionisation)
Total number of particles `=1 - alpha + alpha + alpha = 1 + alpha`
`i = 1 + alpha`
`1.92 = 1+ alpha`
So, `alpha = 1.92 -1 =0.92`
i.e., 9% dissociated.