Question

i ) Resistance of a conductivity cell filled with 0.1 mol/L KCl solution is 100 ohm. If the resistance of the
same cell when filled with 0.02 mol/L KCl solution is 520 ohm , calculate the conductivity and molar
conductivity of 0.02 mol/L KCl solution. The conductivity of 0.1 mol/L KCl solution is 1.29 S/m.
ii ) write the expression for Nernst equation to determine the cell potential .
iii ) how much charge is required for 1 mol of Al^ +3 to Al ?

Answer 1

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Specific conductivity of 0.1 M KCl = 1.29 * 10⁻² S/cm

Let Volume of the Cell = V cm³

Conductivity of 0.01 KCl solution = σ1

= 1.29 * 10⁻² * V S-cm²

Conductivity of 0.052 MX solution = σ2

Conductance of 0.1M KCl solution = G1 = 1/Resistance = 1/85 mho

conductance of 0.052M MX solution = G2 = 1/96 mho

Formula for Conductivity σ = k G, k = cell constant

σ2 / σ1 = G2 / G1

σ2 = G2 * σ1 / G1

= 0.0129 *V* 85/96 = 0.011421875 * V S cm²

Molar conductivity of MX electrolyte:

Λ_m = 1000 * σ2 / c

= 1000 * 0.011421875 *V / (0.052 * V) S-cm²/mole

= 219.65 S-cm²/mole

Answer 2

$\huge\mathfrak\pink{Answer!}$

Specific conductivity of 0.1 M KCl = 1.29 * 10⁻² S/cm

Let Volume of the Cell = V cm³

Conductivity of 0.01 KCl solution = σ1

= 1.29 * 10⁻² * V S-cm²

Conductivity of 0.052 MX solution = σ2

Conductance of 0.1M KCl solution = G1 = 1/Resistance = 1/85 mho

conductance of 0.052M MX solution = G2 = 1/96 mho

Formula for Conductivity σ = k G, k = cell constant

σ2 / σ1 = G2 / G1

σ2 = G2 * σ1 / G1

= 0.0129 *V* 85/96 = 0.011421875 * V S cm²

Molar conductivity of MX electrolyte:

Λ_m = 1000 * σ2 / c

= 1000 * 0.011421875 *V / (0.052 * V) S-cm²/mole

= 219.65 S-cm²/mole

$Hope it helps you$

$ItzRiya$