At 298 K, the specific conductance ( ) of a 0.0020 M NaCl solution is 2.50 10⁻⁴ S cm ⁻¹ . Calculate the molar conductivity ( _m ) of the solution in S cm ^2 mol ⁻¹ .

Step-by-step solution

Given: = 2.50 10⁻⁴ S cm ⁻¹ C = 0.0020 M The molar conductivity _m is given by the formula: _m = 1000 C Substituting the given values: _m = 2.50 10⁻⁴ 1000 0.0020 _m = 0.250 0.0020 _m = 125 S cm ^2 mol ⁻¹ Answer: 125