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]`

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) )

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

0.1mole aqueous solution of NaBr freezes at - 0.335^@C at atmospheric pressure , k_f for water is 1.86^@C . The percentage of dissociation of the salt in solution is