A conductivity cell is filled with a 0.01 M solution of a weak acid. The resistance of the solution is 2500 and the cell constant is 0.5 cm ⁻¹ . If the limiting molar conductivity of the weak acid is 400 S cm ^2 mol ⁻¹ , what is its degree of dissociation?

Step-by-step solution

First, calculate the conductivity ( ) of the solution using the cell constant ( G^* ) and resistance ( R ): = G^* R = 0.5 2500 = 2 10⁻⁴ S cm ⁻¹ Next, calculate the molar conductivity ( _m ) of the solution: _m = 1000 c = 2 10⁻⁴ 1000 0.01 = 0.2 0.01 = 20 S cm ^2 mol ⁻¹ Finally, calculate the degree of dissociation ( ): = _m _m^ = 20 400 = 0.05 Answer: 0.05