Question

The molar conductivity of CH₃COOH at infinite dilution is found to be 387 ohm⁻¹ cm² mol⁻¹ but at a dilution of 1 g mol in 1000 litres it is 55 ohm⁻¹ cm² mol⁻¹. What is the percent dissociation of the acid at the given dilution?

Answer 1

Given: `Lambda_m^0` = 387 Ω⁻¹ cm² mol⁻¹

Λ_m = 55 Ω⁻¹ cm² mol⁻¹

To calculate the percent dissociation of acetic acid, use the formula:

α = `(Lambda_m)/(Lambda_m^0)`    ...(i)

% dissociation = α × 100    ...(ii)

α = `55/387`    ...[By using eqn. (i)]

α = 0.1421

% dissociation = 0.1421 × 100

% dissociation = 14.21%

∴ The percent dissociation of the acid at the given dilution is 14.21%.