Kohlrausch examined the limiting molar conductivity (Λ°m) of a number of strong electrolytes and observed certain regularities. On the basis of these observations he enunciated the Kohlrausch law of independent migration of ions. The law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte. Thus, for sodium chloride:
\(\Lambda^{\circ}_m(\text{NaCl}) = \lambda^{\circ}_{\text{Na}^+} + \lambda^{\circ}_{\text{Cl}^-}\)
Using the Kohlrausch law, it is possible to calculate Λ°m for any electrolyte from the λ° of its individual ions. For a weak electrolyte such as acetic acid, Λ°m can also be obtained from the Λ°m values of strong electrolytes:
\(\Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = \Lambda^{\circ}_m(\text{CH}_3\text{COONa}) + \Lambda^{\circ}_m(\text{HCl}) - \Lambda^{\circ}_m(\text{NaCl})\)
Electrolytes are classified by the charges on the cation and the anion produced on their dissociation in solution. Match List-I with List-II.
| List-I (Electrolyte) | List-II (Type of electrolyte) |
|---|---|
| (A) \( \ce{MgCl2} \) | (I) 1-1 |
| (B) \( \ce{K2SO4} \) | (II) 2-1 |
| (C) \( \ce{CuSO4} \) | (III) 1-2 |
| (D) \( \ce{KBr} \) | (IV) 2-2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-III, C-IV, D-I
The type of an electrolyte is written as two numbers: the charge on the cation first, then the charge on the anion. So NaCl is a 1-1, CaCl2 a 2-1 and MgSO4 a 2-2 electrolyte.
(A) \( \ce{MgCl2} \) — This matches (II) 2-1. It gives \( \ce{Mg^2+} \) and \( \ce{Cl^-} \) ions: cation charge 2, anion charge 1, like CaCl2.
(B) \( \ce{K2SO4} \) — This matches (III) 1-2. It gives \( \ce{K+} \) and \( \ce{SO4^2-} \) ions: cation charge 1, anion charge 2. The order of the numbers matters, so it is not a 2-1 electrolyte.
(C) \( \ce{CuSO4} \) — This matches (IV) 2-2. It gives \( \ce{Cu^2+} \) and \( \ce{SO4^2-} \) ions, both with charge 2, like MgSO4.
(D) \( \ce{KBr} \) — This matches (I) 1-1. It gives \( \ce{K+} \) and \( \ce{Br^-} \) ions, both with charge 1, like NaCl.
NCERT (2026-27 edition):
2.4.2 Variation of Conductivity and Molar Conductivity with Concentration – “…depends on the type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in the solution. Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2 electrolytes respectively.”