Question

The conductivity of 0.20 mol L⁻¹ solution of KCl is 2.48 × 10⁻² S cm⁻¹. Calculate its molar conductivity and degree of dissociation (α). Given λ⁰ (K⁺) = 73.5 S cm² mol⁻¹ and λ⁰ (C1⁻) = 76.5 S cm² mol⁻¹.

Answer 1

Conductivity of KCl solution = 2.48×10^–2 S cm^–1

Concentration of KCl solution = 0.20 mol L^–1

                                          = 0.20 × 1000 mol cm^–3

                                          = 200 mol cm^–3

Molar conductivity

`Lambda_m=k/c`

     `=(2.48xx10^(-2)" S "cm^(-1))/(200" mol "cm^(-3))`

     `=124xx10^(-6)" S "cm^2" mol"^(-1)`

Given:

`lambda_(K^+)^@=73.5" S "cm^2" mol"^(-1)`

`lambda_(Cl^-)^@=76.5" S "cm^2" mol"^(-1)`

Degree of dissociation

`alpha=Lambda_m/Lambda_m^@" "" "" "" ".......("i")`

`Lambda_(m(KCl))^@=lambda_(K+)^@+lambda_(Cl^-)^@`

            `=73.5+76.5`

            `=150" S "cm^2" mol"^(-1)`

`"Substituting the values of "Lambda_m" and "Lambda_(m(KCl))^@" in (i), we get"`

`alpha=(124xx10^(-6)" S "cm^2" mol"^(-1))/(150" S "cm^2" mol"^(-1))=0.82xx10^(-6)`