When a certain conductivity cell was filled with 0.1M KCl , it has a resistance of 85 ohm at 25C. When the same cell was filled with an aqueous solution of 0.052M unknown electrolyte the resistance was 96 ohm. Calculate the molar conductivity of unknown electrolyte at this concentration. (Specific conductivity of 0.1M KCl = 1.29 * 10 -2 ohm/cm)
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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
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
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