A solution containing `0.85 g` of `ZnCI_(2)` in `125.0g` of water freezes at `-0.23^(@)C`. The apparent degree of dissociation of the salt is:
`(k_(f)` for water `= 1.86 K kg mol^(-1)`, atomic mass, `Zn = 65.3` and `CI = 35.5)`

A solution containing 0.85 g of ZnCl_(2) in 125.0g of water freezes at -0.23^(@)C . The apparent degree of dissociation of the salt is: (k_(f) for water = 1.86 K kg mol^(-1) , atomic mass, Zn = 65.3 and Cl = 35.5)

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]

0.5% aqueous KCl has freezing point -0.24^@C . Calcualte the degree of dissociation, If K_f for water is 1.86 K kg mol^(-1) .

A solution containing 1g of sodium chloride in 100g of water freezes at - 0.604^@C . Calculate the degree of dissociation of sodium chloride. (Na = 23, Cl = 35.5, Kf for water = 1.87 k "mol"^(-1) )

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) )

The freezing point of a solution containing 5.85 g of NaCl in 100 8 of water is -3.348^(@)C . Calculate van't Hoff factor for this solution. What will be the experimental molecular weight of NaCl? ( K_(f) for water = 1.86 K kg mol^(-1) , at. wt. Na = 23, Cl = 35.5)