A solution containing 0.5 g of KCI dissolved in 100 g of water and freezes at `-0.24^(@)C`. Calculate the degree of dissociation of the salt. (`K_(f)` for water =`1.86^(@)C`. [Atomic weight K = 39, Cl = 35.5]

Explanation

Observed molecular mass =` M= (K_(f) xx w xx 1000)/(DeltaT_(f) xx W)= (1.86 xx 0.5 xx 1000)/(0.24 xx 100)= 38.75`
Normal molecular mass of KCl=74.5
Van.t Hoff factor=`("Normal molar mass")/("Observed molar mass")= (74.5)/(38.75)= 1.92`
KCI dissociates as `KCl to K^(+)+Cl^(-)`
Moles after dissociation `(1-alpha) alpha alpha`
Total no. of moles after dissociation=`1+alpha`
`i= ("Observed moles of solute")/("Normal moles of solute")= (1+alpha)/(1)= (1+alpha)/(1)= 1.92`
`alpha= 1.92-1= 0.92 `
Degree of dissociation = 92%