AIIMS2016ChemistryElectrochemistry
A conductivity cell has a cell constant of 0.5 ~cm ⁻¹ . This cell when filled with 0.01 M NaCl solution has a resistance of 384 ohms at 25^ C . Calculate the equivalent conductance of the given solution.
Options
- A130.2 ⁻¹ ~cm ^2( ~g eq )⁻¹
- B137.4 ⁻¹ ~cm ^2( ~g eq )⁻¹
- C154.6 ⁻¹ ~cm ^2( ~g eq )⁻¹
- D169.2 ⁻¹ ~cm ^2( ~g eq )⁻¹
Correct answer
A. 130.2 ⁻¹ ~cm ^2( ~g eq )⁻¹
Step-by-step solution
aligned & Equivalent conductance, _ e q = 1000 Normality & where, (Specific conductance) =C l a & = 1 R l a = 1 384 0.5 & =1.302 10⁻³ ohm ⁻¹ ~cm ⁻¹ & _ e q = 1.302 10⁻³ 1000 0.01 & =130.2 ohm ⁻¹ ~cm ^2( ~g eq )⁻¹ aligned