If limiting molar conductance of \(NaCl,\, \ K_3PO_4\) and \(KCl\) are \(110.2,\, \ 98.8\) and \(134.6\, \ \Omega ^{-1}cm^2mol^{-1}\) respectively, the limiting molar conductance of \(Na_3PO_4\) will be : \((in\, \ \Omega ^{-1}cm^2mol^{-1})\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 25.6
By Kohlrausch’s law, the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions. So Λ°m(Na3PO4) can be built from the given salts:
Λ°(\( \ce{Na3PO4} \)) = 3λ°(\( \ce{Na^{+}} \)) + λ°(\( \ce{PO4^{3-}} \))
= 3Λ°(NaCl) + Λ°(\( \ce{K3PO4} \)) − 3Λ°(KCl)
= 3(110.2) + 98.8 − 3(134.6)
= 330.6 + 98.8 − 403.8
= 25.6 Ω−1 cm2 mol−1
NCERT (2026-27 edition):
Kohlrausch Law of Independent Migration of Ions – “… The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.”