Question

Read the passage carefully and answer the questions that follow.

Vishal set up an experiment to find the resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with A.C. power in the audio frequency range of 550 to 5000 cycles per second. Once the resistance was calculated from the null point, he also calculated the conductivity (κ) and molar conductivity (Λm) and recorded his readings in tabular form, which is given below. S. No. Conc. (M) κ (S cm−1) Λm (S cm2 mol−1) 1. 1.00 111.3 × 10−3 1113 2. 0.10 12.9 × 10−3 129.0 3. 0.01 1.41 × 10−3 141.0

1. Why did the molar conductivity increase, though the conductivity decreased with dilution?
2. If molar conductivity at infinite dilution `(Λ_m^o)` of KCl is 150.0 S cm² mol⁻¹, calculate the degree of dissociation of 0.01 M KCl.
3. The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10⁻⁴ S cm⁻¹. Calculate the molar conductivity of the solution.

Answer 1

(i) Conductivity (κ) is the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases, so κ decreases.

Molar conductivity (Λ_m) is the conductivity of the solution containing 1 mole of the electrolyte.

Λ_m = κ × V_m 

On dilution, V_m​ (volume containing 1 mole) increases and inter-ionic attraction decreases, so ionic mobility increases. Hence, Λ_m increases on dilution.

(ii) Given: `Λ_m^o` = 150.0 S cm² mol⁻¹

For 0.01 M KCl, Λ_m = 141.0 S cm² mol⁻¹

Degree of dissociation (α) = `Λ_m/Λ_m^o`

= `141.0/150.0`

= 0.94 or 94%

(iii) Given: κ = 1.65 × 10⁻⁴ S cm⁻¹

Molarity = 0.01 M

`Λ_m = (κ xx 1000)/"Molarity"` 

= `(1.65 xx 10^(-4) xx 1000)/0.01`

= 16.5 S cm² mol⁻¹