JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

A $5.0m\mathrm{mol}{\mathrm{dm}}^{-3}$ aqueous solution of $\mathrm{KCl}$ has a conductance of $0.55\mathrm{mS}$ when measured in a cell constant $1.3{\mathrm{cm}}^{-1}$ . The molar conductivity of this solution is _______ ${\mathrm{mSm}}^{2}{\mathrm{mol}}^{-1}$ . (Round off to the Nearest Integer)

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

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Solution

Given ${\mathrm{conc}}^{n}$ of $\mathrm{KCl}=\frac{m.\mathrm{mol}}{L}$

: Conductance $(G)=0.55\mathrm{mS}$

: Cell constant $(\frac{\ell }{A})=1.3{\mathrm{cm}}^{-1}$

To Calculate : Molar conductivity $({\lambda }_{m})$ of sol.

$\rightarrow$ Since ${\Lambda }_{m}=\frac{1}{1000}\times \frac{k}{M}$ $...(1)$

$\rightarrow$ Molarity $=5\times {10}^{-3}\frac{\mathrm{mol}}{L}$

$\rightarrow$ Conductivity $=G\times (\frac{\ell }{A})=0.55\mathrm{mS}\times \frac{1.3}{\frac{1}{100}}{m}^{-1}$

$=55\times 1.3{\mathrm{mSm}}^{-1}$

${\mathrm{eq}}^{n}(1)$ ${\Lambda }_{m}=\frac{1}{1000}\times \frac{55\times 1.3}{(\frac{5}{1000})}\frac{{\mathrm{mSm}}^{2}}{\mathrm{mol}}$

$\Rightarrow {\Lambda }_{m}=14.3\frac{{\mathrm{mSm}}^{2}}{\mathrm{mol}}$

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JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

Given ${\mathrm{conc}}^{n}$ of $\mathrm{KCl}=\frac{m.\mathrm{mol}}{L}$

: Conductance $(G)=0.55\mathrm{mS}$

: Cell constant $(\frac{\ell }{A})=1.3{\mathrm{cm}}^{-1}$

To Calculate : Molar conductivity $({\lambda }_{m})$ of sol.

$\rightarrow$ Since ${\Lambda }_{m}=\frac{1}{1000}\times \frac{k}{M}$ $...(1)$

$\rightarrow$ Molarity $=5\times {10}^{-3}\frac{\mathrm{mol}}{L}$

$\rightarrow$ Conductivity $=G\times (\frac{\ell }{A})=0.55\mathrm{mS}\times \frac{1.3}{\frac{1}{100}}{m}^{-1}$

$=55\times 1.3{\mathrm{mSm}}^{-1}$

${\mathrm{eq}}^{n}(1)$ ${\Lambda }_{m}=\frac{1}{1000}\times \frac{55\times 1.3}{(\frac{5}{1000})}\frac{{\mathrm{mSm}}^{2}}{\mathrm{mol}}$

$\Rightarrow {\Lambda }_{m}=14.3\frac{{\mathrm{mSm}}^{2}}{\mathrm{mol}}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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JEE Main 2021 — Chemistry Physical Chemistry

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Solution

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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