Exams › JEE Main › Chemistry

For a strong electrolyte AB in water at 25 deg C, the molar conductivity Lambdaₘ varies with concentration as: at C = 0.25 mol/L, Lambdaₘ = 160 S cm²/mol; at C = 1 mol/L, Lambdaₘ = 150 S cm²/mol. Using the Debye-Huckel-Onsager (Kohlrausch) relation, find the limiting molar conductivity Lambda°m (in S cm² mol⁻¹).

1. 160
2. 150
3. 170
4. 180

Correct answer: 170

Solution

Kohlrausch's law for strong electrolytes: Lambdaₘ = Lambda°m - K*sqrt(C). At C = 0.25, sqrt(C) = 0.5: 160 = Lambda°m - 0.5K. At C = 1, sqrt(C) = 1: 150 = Lambda°m - K. Subtracting: 160 - 150 = (Lambda°m - 0.5K) - (Lambda°m - K) => 10 = 0.5K => K = 20. Then Lambda°m = 150 + K = 150 + 20 = 170 S cm² mol⁻¹.

Related JEE Main Chemistry questions

• During electrolysis of an aqueous solution of dipotassium succinate, which gas is evolved at the electrodes?
• Which expression gives the equivalent conductance at infinite dilution of Al2(SO4)3, if Λ°Al3+ and Λ°SO4²− denote the equivalent conductances at infinite dilution of the corresponding ions?
• The standard emf of a Daniell cell is 1.10 V. What is the greatest amount of electrical work that can be extracted from this cell?
• During the recharging of a lead storage battery, which reaction takes place at the cathode?
• The limiting molar ionic conductivities of the ions X2+ and Y2− are 57 and 73, respectively. What will be the molar conductivity of the electrolyte formed by these ions?
• According to Kohlrausch’s law, at what condition does each ion contribute a fixed amount to the conductance of an electrolyte, irrespective of the other ion present?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →