`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`.

0.5g KCl was dissolved in 100g water and the solution originally at 20^@C , freeze at -0.24^@C . Calculate the percentage ionization of salt, K_f for per 1000g of water = 1.86K .

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]

1 . 5 of Ba (NO_(3))_(2) dissolved in 100 g of water shows a depression in freezing point equal to 0 . 28^(@)C . What is the percentage dissociation of the salt ? ( K_(f) for water = 1 . 86 K/m and molar mass of Ba (NO_(3))_(2) = 261.)

When 0.6 g of urea is dissolved in 100 g of water, the water will boil at ( K_(b) for water = 0.52 Km^(-1) and normal boiling point of water = 100^(@)C ) :

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - 0.34^(@) C. what is molar mass of the substance ? (K_(f)" for water " = 1.86 k kg " mol"^(-1))