Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is $100 \Omega$. The conductivity of this solution is $1.29 \mathrm{~S} \mathrm{~m}^{-1}$. Resistance of the same cell when filled with 0.02 M of the same solution is $520 \Omega$. The molar conductivity of 0.02 M solution of the electrolyte will be

(A) $124 \times 10^{-4} \mathrm{~s} \mathrm{~m}^{2} \mathrm{~mol}^{-1}$

(B) $1240 \times 10^{-4} \mathrm{~s} \mathrm{~m}^{2} \mathrm{~mol}^{-1}$

(C) $1.24 \times 10^{-4} \mathrm{~s} \mathrm{~m}^{2} \mathrm{~mol}^{-1}$

(D) $12.4 \times 10^{-4} \mathrm{~s} \mathrm{~m}^{2} \mathrm{~mol}^{-1}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Redox Reactions and Electrochemistry

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is

100Ω

. The conductivity of this solution is

1.29 S m^{- 1}

. Resistance of the same cell when filled with 0.02 M of the same solution is

520Ω

. The molar conductivity of 0.02 M solution of the electrolyte will be

(A) $124 \times 1 0^{- 4} s m^{2} mol^{- 1}$ ✓ Correct
(B) $1240 \times 1 0^{- 4} s m^{2} mol^{- 1}$
(C) $1.24 \times 1 0^{- 4} s m^{2} mol^{- 1}$
(D) $12.4 \times 1 0^{- 4} s m^{2} mol^{- 1}$

Answer: (A)

Solution

From the 0.1 M data the cell constant is G* = κ·R = 1.29 × 100 = 129 m⁻¹. For the 0.02 M solution, κ = G*/R = 129/520 = 0.248 S m⁻¹. Molar conductivity Λ_m = κ/c with c = 0.02 mol L⁻¹ = 20 mol m⁻³: Λ_m = 0.248/20 = 0.0124 = 124 × 10⁻⁴ S m² mol⁻¹ → option A. The printed paper labelled the second fill as 0.2 M while asking for the 0.02 M value (internally inconsistent); this is the standard NCERT problem whose second fill is 0.02 M, so the stem is corrected to 0.02 M and the self-consistent answer is 124 × 10⁻⁴.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Redox Reactions and Electrochemistry

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