A `0.01 m` aqueous solution of potassium ferricy freezes at `-0.062^(@)C`. What is the apparent
Channel: Doubtnut
Date: 2020-08-23

A `0.01 m` aqueous solution of potassium ferricy freezes at `-0.062^(@)C`. What is the apparent percentage of dissociation? `(K_(3)` for `H_(2)O -1.86)`.  Strategy: Potassium is `K_(3)[Fc(CN)_(6)]`. The appearent percentage of dissociation is degree of dissociation x 100%. To calculate degree of dissociation we must determine the value of van't hoff factor for which we need to work out calculated colligative property.