If an electrolyte on dissociation gives \( v_+ \) cations and \( v_- \) anions, then its limiting molar conductivity is given by
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\Lambda_m^0 = v_+\lambda^0_+ + v_-\lambda^0_-\)
Kohlrausch’s law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cations and anions. At infinite dilution each ion moves on its own and adds a fixed amount, \( \lambda^{0} \), to the conductivity.
If one formula unit gives \( \nu_+ \) cations and \( \nu_- \) anions, each cation contributes \( \lambda^{0}_{+} \) and each anion \( \lambda^{0}_{-} \):
\( \displaystyle \Lambda^{0}_{m} = \nu_+\lambda^{0}_{+} + \nu_-\lambda^{0}_{-} \)
For example, calcium chloride gives one \( \ce{Ca^2+} \) and two \( \ce{Cl-} \) ions:
\( \displaystyle \ce{CaCl2 -> Ca^2+ + 2Cl^-} \)
\( \displaystyle \Lambda^{0}_{m}(\ce{CaCl2}) = \lambda^{0}(\ce{Ca^2+}) + 2\lambda^{0}(\ce{Cl-}) \)
Shortcut: The contributions of the ions add up, so Options (1) and (2), with a minus sign, are out. Checking with \( \ce{CaCl2} \), the factor 2 must go with \( \lambda^{0}(\ce{Cl-}) \), the anion, which leaves Option (4).
Why the other options are incorrect
Option (1) \(\Lambda_m^0 = v_-\lambda^0_+ - v_+\lambda^0_-\) — Incorrect. The contributions are subtracted, and each ion is multiplied by the number of the other ion.
Option (2) \(\Lambda_m^0 = v_+\lambda^0_+ - v_-\lambda^0_-\) — Incorrect. The numbers are paired correctly, but the anion contribution is subtracted; every ion carries current and adds to the conductivity.
Option (3) \(\Lambda_m^0 = v_-\lambda^0_+ + v_+\lambda^0_-\) — Incorrect. The numbers are swapped; for \( \ce{CaCl2} \) it would give \( 2\lambda^{0}(\ce{Ca^2+}) + \lambda^{0}(\ce{Cl-}) \), which counts two calcium ions.
NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – …
