Question

0.1 M KCl solution offered a resistance of 100 ohms in a conductivity cell at 298 K. If the cell constant of the cell is 1.29 cm⁻¹, calculate the molar conductivity of KCl solution.

Answer 1

We know , molar conductivity increases when we dilute the solution. As on dilution interiomjc interactions are overcome and ions are free to move.

Given, Cell constant of the cell is 1.29 cm⁻¹

concentration of solution is 0.1 M

resistance in the conductivity cell is 100Ω

now, use formula, cell constant = κ × resistance

κ = cell constant/resistance = 1.29 cm⁻¹/100Ω = 0.0129 Scm⁻¹

         Now use formula, $\Lambda_m=\frac{1000\times\kappa}{M}$

So, $\Lambda_m$ = 1000 × 0.0129/0.1 = 129 Scm²mol⁻¹

Answer 2

0.1 M KCl solution offered a resistance of 100 ohms in a conductivity cell at 298 K.The cell constant of the cell is 1.29 cm⁻.

We know , molar equivalent conductivity increases when we dilute the solution based on oswald's dilution law. As on dilution inter ionic interactions are overcome and ions are free to move.

Given, Cell constant of the cell is 1.29 cm⁻¹

Concentration of solution = 0.1 M

resistance in the conductivity cell = 100Ω

now, use formula, cell constant = κ × resistance

κ = cell constant/resistance = 1.29 cm⁻¹/100Ω = 0.0129 Scm⁻¹

Now using the formula,

So,   κ = 1000 × 0.0129/0.1 = 129 Scm²mol⁻¹