The limiting molar conductivities ( _m^0 ) for HCl , NaCl and CH ₃ COONa are 426 S cm ^2 mol ⁻¹ , 126 S cm ^2 mol ⁻¹ and 90 S cm ^2 mol ⁻¹ respectively. If the molar conductivity of a 0.001 M CH ₃ COOH solution is 39 S cm ^2 mol ⁻¹ , the degree of dissociation of acetic acid at this concentration is:

Step-by-step solution

According to Kohlrausch's law, the limiting molar conductivity of acetic acid ( CH ₃ COOH ) is calculated from the given strong electrolytes as follows: _m^0( CH ₃ COOH ) = _m^0( CH ₃ COONa ) + _m^0( HCl ) - _m^0( NaCl ) _m^0( CH ₃ COOH ) = 90 + 426 - 126 = 390 S cm ^2 mol ⁻¹ The degree of dissociation ( ) is given by the ratio of molar conductivity at a given concentration ( _m ) to the limiting molar conductivity ( _m^0 ): = _m _m^0 = 39 390 = 0.1 The correct degree of dissociation is 0.1 . The distractor 0.01 ar