Resistance of 0.2 M solution of an electrolyte is 50 . The conductivity of the solution is 1.3 Sm ⁻¹ . If the resistance of 0.4 M solution of the same electrolyte is 260 , its molar conductivity is:

Step-by-step solution

The cell constant G^* is given by the product of conductivity and resistance of the first solution: G^* = ₁ R₁ G^* = 1.3 S m⁻¹ 50 = 65 m⁻¹ The conductivity of the 0.4 M solution, ₂ , is: ₂ = G^* R₂ = 65 m⁻¹ 260 = 0.25 S m⁻¹ The concentration of the second solution in SI units is: C₂ = 0.4 mol L⁻¹ = 0.4 10^3 mol m⁻³ = 400 mol m⁻³ The molar conductivity _m is given by: _m = ₂ C₂ _m = 0.25 S m⁻¹ 400 mol m⁻³ = 6.25 10⁻⁴ S m^2 mol⁻¹ Answer: 6.25 10⁻⁴ Sm^2mol⁻¹