Question

1 gm of a monobasic acid in 100 gm of water lowers the freezing point by 0.168 C . If 0.2 gm of the same acid require 15 ml of 1/10N alkali for complete neutralization, the degree of dissociation of acid is if Kf for water=1.86 K/mol

Answer 1

Hey dear,

● Answer -
α = 0.66

● Explanation -
For monobasic acid, normality is given by -
N = n / V × 1000 = 1000W / MV
M = 1000W / NV
M = 1000 × 0.25 / (0.1 × 15)
M = 166.6 g

Molality is calculated by -
m = moles of solute /volume of solvent × 1000
m = (1/166.6) × 1000 / 100
m = 0.06 mol/L

Formula for depression of freezing point is -
∆Tf = Kf.m
∆Tf = 1.86 × 0.06
∆Tf = 0.1116

Degree of dissociation is calculated by -
α = ∆Tf(obs) / ∆Tf - 1
α = 0.186 / 0.116 - 1
α = 1.66 - 1
α = 0.66

Hope this helps you...