Exams › JEE Main › Chemistry

At 25°C, the limiting molar conductances in water are given below for these electrolytes: KCl = 149.9 S cm² mol−1, KNO3 = 145 S cm² mol−1, HCl = 426.2 S cm² mol−1, NaOAc = 91 S cm² mol−1, and NaCl = 126.5 S cm² mol−1. Using these values, determine the limiting molar conductance of HOAc in water.

1. 217.5
2. 390.7
3. 552.7
4. 517.2

Correct answer: 390.7

Solution

The limiting molar conductance of acetic acid (HOAc) can be calculated using the known conductances of its ions and the principle of additivity for strong electrolytes. By combining the contributions from acetate (NaOAc) and the dissociated protons from HCl, we arrive at the correct value of 390.7 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 →