An aqueous solution of a salt M(NO₃)₂ has a molality of 0.1 m and freezes at -0.4464^ C . If the molal depression constant for water is 1.86 K kg mol ⁻¹ , the percentage dissociation of the salt is ______ .

Step-by-step solution

The depression in freezing point is given by: T_f = T_f^ - T_f = 0 - (-0.4464) = 0.4464 K Using the formula for freezing point depression: T_f = i K_f m 0.4464 = i 1.86 0.1 0.4464 = 0.186 i i = 2.4 For the salt M(NO₃)₂ , the number of ions produced per formula unit is n = 3 . The relation between van't Hoff factor ( i ) and degree of dissociation ( ) is: i = 1 + (n - 1) 2.4 = 1 + (3 - 1) 1.4 = 2 = 0.7 The percentage dissociation is 0.7 100 = 70 . Answer: 70