0.01 molal aqueous solution of `K_(3)[Fe(CN)_(6)]` freezes at `-0.062^(@)C`. Calculate percentage dissociation `(k_(f)=1.86)`

0.01 m aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C . What is the apparent percentage of dissociation ? ( K_(f) for water = 1.86" K kg mol"^(-1) )

A 0.01m aqueous solution of K_(3)[Fe(CN)_(6)] freezes ar -0.062^(@)C . What is the apparent percentage of dissociation? [K_(f) for water = 1.86]

A 0.01m aqueous solution of K_(3)[Fe(CN)_(6)] freezes ar -0.062^(@)C . What is the apparent percentage of dissociation? [K_(f) for water = 1.86]

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.

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .

A 0.01 molal solution of ammonia freezes at -0.02^(@)C . Calculate the van't Hoff factor, i and the percentage dissociation of ammonia in water. (K_(f(H_(2O))))=1.86 deg "molal"^(-1) .