Question

when 0.1 mol ch3cooh is present in a litre solvent it shows elevation in bolling point of 0.75c. acid dissociation constant will be (kb=5k kg mol^-1)​

Answer 1

your question -> when 0.1m ch3cooh is present in a litre solvent it shows elevation in bolling point of 0.75°c. acid dissociation constant will be (kb=5k kg mol^-1)

solution : elevation in boiling point, ∆Tb = 0.75°C

molality of CH3COOH , m = 0.1molal

dissociation of CH3COOH,

CH3COOH ⇔ CH3COO- + H^+

so, Van't Hoff's factor, i = 1 - α + 2α

= 1 + α

using formula, ∆Tb = i × m × Kb

or, 0.75°C = (1 + α) × 0.1molal × 5K kg/mol

or, 1 + α = (0.75)/(0.5) = 1.5

so, α = 0.5

hence, degree of dissociation or dissociation constant, α = 0.5