The limiting molar conductivities $\Lambda^{\circ}$ for $\mathrm{NaCl}, \mathrm{KBr}$ and KCl are 126, 152 and $150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. The $\Lambda^{\circ}$ for NaBr is

(A) $128 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

(B) $302 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

(C) $278 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

(D) $176 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

JEE Main · 2004 · 24 Apr 2004 · Offline · Redox Reactions and Electrochemistry

The limiting molar conductivities

Λ^{\circ}

for

NaCl, KBr

and KCl are 126, 152 and

150 S cm^{2} mol^{- 1}

respectively. The

Λ^{\circ}

for NaBr is

Answer: (A)

Kohlrausch's law lets each molar conductivity be split into independent ionic parts. Writing

Λ_{NaCl}^{\circ} = λ_{Na X^{+}} + λ_{Cl X^{-}} = 126

Λ_{KBr}^{\circ} = λ_{K X^{+}} + λ_{Br X^{-}} = 152

and

Λ_{KCl}^{\circ} = λ_{K X^{+}} + λ_{Cl X^{-}} = 150

, the value for

NaBr

follows from adding the first two and subtracting the third:

Λ_{NaBr}^{\circ} = 126 + 152 - 150 = 128 S cm^{2} mol^{- 1}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option