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Chemistry · Electrochemistry

JEE Main 2026 — 21 January, Morning Shift — Question 64

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 (a) << 1)

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

pH=5\mathrm{pH}=5 [H+]=10−5=[HX].α\left[\mathrm{H}^{+}\right]=10^{-5}=[\mathrm{HX}] . \alpha =[HX]⋅ΛmΛm∞=[\mathrm{HX}] \cdot \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\infty}} Λm=k×1000[HX]\Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{[\mathrm{HX}]} K=G.G∗=4×10−5×151=6×10−4 S.cm−1\mathrm{K}=\mathrm{G} . \mathrm{G}^{*}=4 \times 10^{-5} \times \frac{15}{1}=6 \times 10^{-4} \mathrm{~S} . \mathrm{cm}^{-1} [H+]=10−5=[HX]×6×10−4×1000Λm∞×[HX]\left[\mathrm{H}^{+}\right]=10^{-5}=[\mathrm{HX}] \times \frac{6 \times 10^{-4} \times 1000}{\Lambda_{\mathrm{m}}^{\infty} \times[\mathrm{HX}]} Λm∞=60000S.cm2 mol−1\Lambda_{\mathrm{m}}^{\infty}=60000 \mathrm{S.cm}^{2} \mathrm{~mol}^{-1} Λm∞=6 S.m2 mol−1\Lambda_{\mathrm{m}}^{\infty}=6 \mathrm{~S} . \mathrm{m}^{2} \mathrm{~mol}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law
The pH and conductance of a weak acid (HX) was found to be 5 and 4 ×… | JEE Main 2026 PYQ with Solution · DhiX AI