Solved Examples on Conductance of Electrolytic Solutions
Example 1: The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500 ohm in a conductivity cell. If the cell constant of the cell is 0.367 per cm, calculate the molar conductivity of the solution.
Solution:
Cell constant = Conductivity/Conductance = Conductivity × Resistance
= (0.146 × 10-3) Scm-1 × 1500 ohm
= 0.219cm-1
Example 2: The Conductivity of 0.20 m solutions of KCl at 298K is 0.0248 Scm-1. Calculate its molar conductivity.
Solution:
Molar conductivity = (κ × 1000)/Molarity = [(0.0248Scm-1 × 1000cm3 L-1)/ 0.20 molL-1]
= 124 Scm2 mol-1
Example 3: The electrical resistance of a column of 0.05mol/L NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate resistivity, conductivity, molar conductivity.
Solution:
Area = πr2 = 3.14 × 0.52 cm = 0.785 cm2 = 0.785 ×10-4 m2, ρ = 5.55 × 103 ohm
R = ρ l/A = [(5.55× 103 ohm × 0.785cm2)/50 cm]
= 87.135 ohm cm
Conductivity = κ = 1/ρ = (1/87.135)Scm-1 = 0.01148 Scm-1
Molar Conductivity = [(κ × 1000)/c] cm3L-1 = (0.01148 Scm-1 × 1000 cm3L-1)/0.05molL-1
= 229.6 Scm2 mol-1
Conductance of Electrolytic Solutions
Electrochemistry is a branch of chemistry, and it deals with the study of the production of electricity from the energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous (requiring the input of external energy) chemical transformation. Electrochemistry deals with how much chemical energy produced in a redox reaction can be converted into electrical energy. A redox reaction is that in which oxidation and reduction take place simultaneously. The arrangements used to bring about the chemical transformations are called electrochemical cells. The cells are used to convert chemical energy into electrical energy and electrical energy into chemical energy.