The limiting molar conductivities of HCOONa , HCl and NaCl are 100 , 425 and 125 S cm ^2 mol ⁻¹ respectively. If the molar conductivity of a 0.01 M solution of HCOOH is 40 S cm ^2 mol ⁻¹ , what is its degree of dissociation?

Step-by-step solution

According to Kohlrausch's law, the limiting molar conductivity of HCOOH is calculated from the strong electrolytes as follows: ^ _m( HCOOH ) = ^ _m( HCOONa ) + ^ _m( HCl ) - ^ _m( NaCl ) ^ _m( HCOOH ) = 100 + 425 - 125 = 400 S cm ^2 mol ⁻¹ The degree of dissociation ( ) is given by the ratio of molar conductivity at a given concentration to the limiting molar conductivity: = _m ^ _m = 40 400 = 0.1 Answer: 0.1