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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2026 · 21 January · Morning Shift · Chemistry · Electrochemistry

The pH and conductance of a weak acid (HX) was found to be 5 and 4...

The pH and conductance of a weak acid (HX) was found to be 5 and 4×10−5 S4 \times 10^{-5} \mathrm{~S}, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm21 \mathrm{~cm}^{2} were at a distance of 15 cm apart. The value of the limiting molar conductivity is ____\_\_\_\_ Sm2 mol−1\mathrm{S} \mathrm{m}^{2} \mathrm{~mol}^{-1}. (nearest integer) (Given : degree of dissociation of the weak acid (α)≪1(\alpha) \ll 1)

Correct answer

0

Step-by-step solution

Given: pH = 5, so [H+]=10−5[H^+] = 10^{-5} M; G = 4×10−54 \times 10^{-5} S; A = 1 cm²; l = 15 cm
Cell constant = l/A = 15 cm⁻¹
Specific conductivity: κ=G×(l/A)=4×10−5×15=6×10−4\kappa = G \times (l/A) = 4 \times 10^{-5} \times 15 = 6 \times 10^{-4} S/cm
For weak acid HX: [H+]=Cα=10−5[H^+] = C\alpha = 10^{-5} M
Using κ=Cα×Λ°m1000\kappa = \frac{C\alpha \times \Lambda°_m}{1000}
Λ°m=1000×κCα=1000×6×10−410−5=6×104\Lambda°_m = \frac{1000 \times \kappa}{C\alpha} = \frac{1000 \times 6 \times 10^{-4}}{10^{-5}} = 6 \times 10^4 S cm² mol⁻¹
Converting: Λ°m=6×104×10−4=6\Lambda°_m = 6 \times 10^4 \times 10^{-4} = 6 S m² mol⁻¹
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