If 0.1 molal aqueous solution of `MgSO_(4)` freezes at `-0.225^(@)C`, then degree of dissociation of `MgSO_(4)` is `(K_(f)` for `H_(2)O ` is `1.86Kkgmol^(-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 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 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)

The freezing point of 0.08 molal aq NaHSO_(4) solution is -0.372 . Calculate alpha (degree of dessociation) of HSO_(4)^(-) given alpha (degree of dissociation) of NaHSO_(4) is 100% & K_(f) for H_(2)O=1.86 (K)/(m) . Give answer after multiplying by 10.

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molality of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

A 0.2 molal solution of KCl freezes at -0.68^(@)C . If K_(f) for H_(2)O is 1.86 , the degree of dissociation of KCl is a. 85% , b. 83% , c. 65% , d. 90%

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) .

Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na_(2)SO_(4) . If this solution actually freezes at -0.320 ^(@)C , what would be the value of Van't Hoff factor ? ( K_(f) for water is 1.86 ^(@)C mol^(-1) ) .

1.0 molal aqueous solution of an electrolyte X_(3)Y_(2) is 25% ionized. The boiling point of the solution is (K_(b) for H_(2)O = 0.52 K kg//mol) :