`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

Let us first calculate the observed molar mass as :
`M_(B) = (k_(f) xx w_(B) xx 1000)/(DeltaT_(f) xx w_(A))`
`W_(b) = 0.5 g , w_(A) = 100g , Delta T_(f) = 0- (-0.24) = 0.24^(@),K = 1.86`
`therefore " " M_(B) = (1.86 xx 0.5 xx 1000)/(0.24 xx 100) = 38.75`
Normal molar mass of KCl = 39+35.5=74.5
Van.t Hoff factor ,
`i.=("Normal molar mass")/("observed molar mass ") = (74.5)/(38.75) = 1.92`
KCl ionizes as :
If `alpha ` is the degree of ionziation then .
`{:(,,KCl hArr,K^(+)+,Cl^(-)),(,"Initial moles",1,0,0),(,"Moles after dissoication",1-alpha,alpha,alpha):}`
Total number of moles after dissociation ` = 1- alpha + alpha + alpha `
`i=("Obserbed moles of solute ")/(" Normal moles of solute ") = (1+alpha)/(1)`
`therefore " " (1+alpha)/(1) = 1.92 or 1 + alpha = 1.92`
`or " " alpha = 1.92 -1 = 0.92`
Percentage ionization = 92%.