JEE Main 2022 — Chemistry Physical Chemistry

2022mcqeasy

The molar conductivity of a conductivity cell filled with $10$ moles of $20\mathrm{mL}\mathrm{NaCl}$ solution is ${\Lambda }_{m1}$ and that of $20$ moles another identical cell heaving $80\mathrm{mL}\mathrm{NaCl}$ solution is ${\Lambda }_{m2}$ , The conductivities exhibited by these two cells are same.

The relationship between ${\Lambda }_{m2}$ and ${\Lambda }_{m1}$ is

Official previous-year question

Held on 25 Jul 2022 · Verified 6 Jul 2026.

Options

A
${\Lambda }_{m2}=2{\Lambda }_{m1}$
B
${\Lambda }_{m2}={\Lambda }_{m1}/2$
C
${\Lambda }_{m2}={\Lambda }_{m1}$
D
${\Lambda }_{m2}=4{\Lambda }_{m1}$

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Solution

${\Lambda }_{m}=\kappa \times \frac{1000}{M}$

$\Rightarrow {\Lambda }_{m}\propto \frac{1}{M}$

$\frac{{\Lambda }_{m1}}{{\Lambda }_{m2}}=\frac{{M}_{2}}{{M}_{1}}=\frac{\frac{20}{80}}{\frac{10}{20}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}$

$\Rightarrow {\Lambda }_{m2}=2{\Lambda }_{m1}$

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Solution

${\Lambda }_{m}=\kappa \times \frac{1000}{M}$

$\Rightarrow {\Lambda }_{m}\propto \frac{1}{M}$

$\frac{{\Lambda }_{m1}}{{\Lambda }_{m2}}=\frac{{M}_{2}}{{M}_{1}}=\frac{\frac{20}{80}}{\frac{10}{20}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}$

$\Rightarrow {\Lambda }_{m2}=2{\Lambda }_{m1}$

JEE Main 2022 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2022 — Chemistry Physical Chemistry

Options

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