JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

A $\mathrm{KCl}$ solution of conductivity $0.14S{m}^{-1}$ shows a resistance of $4.19\Omega$ in a conductivity cell. If the same cell is filled with an $\mathrm{HCl}$ solution, the resistance drops to $1.03\Omega$ . The conductivity of the HCl solution is ____ $\times {10}^{-2}S{m}^{-1}$ . (Round off to the Nearest Integer).

Official previous-year question

Held on 17 Mar 2021 · Verified 6 Jul 2026.

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Solution

The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by $\kappa$ , \lambda, or \kappa. Heat transfer occurs at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity

$\kappa =\frac{1}{R}\cdot {G}^{*}$

For same conductivity cell, ${G}^{*}$ is constant and hence $\kappa \cdot R.=$ constant.

$\therefore 0.14\times 4.19=\kappa \times 1.03$

or, $=\frac{0.14\times 4.19}{1.03}$

$=0.5695{\mathrm{Sm}}^{-1}$

$=56.95\times {10}^{-2}{\mathrm{Sm}}^{-1}\approx 57\times {10}^{-2}{\mathrm{Sm}}^{-1}$

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JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

Official previous-year question

Held on 17 Mar 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\kappa =\frac{1}{R}\cdot {G}^{*}$

For same conductivity cell, ${G}^{*}$ is constant and hence $\kappa \cdot R.=$ constant.

$\therefore 0.14\times 4.19=\kappa \times 1.03$

or, $=\frac{0.14\times 4.19}{1.03}$

$=0.5695{\mathrm{Sm}}^{-1}$

$=56.95\times {10}^{-2}{\mathrm{Sm}}^{-1}\approx 57\times {10}^{-2}{\mathrm{Sm}}^{-1}$

JEE Main 2021 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2021 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry