JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

The density of a monobasic strong acid (Molar mass $24.2g\mathrm{mol})$ is $1.21\mathrm{kg}L$ . The volume of its solution required for the complete neutralization of $25\mathrm{mL}$ of $0.24M\mathrm{NaOH}$ is ${10}^{-2}\mathrm{mL}$ (Nearest integer)

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

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Solution

millimole of $\mathrm{NaOH}=0.24\times 25$

For nuetralisation, equal moles of monoacidic base and monobasic acid must be present:

$\therefore$ millimole of acid $=0.24\times 25$

$\Rightarrow$ mass of acid $=0.24\times 25\times 24.2g$

for pure acid,

$V=\frac{w}{d};(d=1.21\mathrm{kg}/L=1.21g/\mathrm{ml})$

$\therefore V=\frac{0.24\times 25\times 24.2}{1.12}\times {10}^{-3}$

$=120\times {10}^{-3}\mathrm{ml}$

$=12\times {10}^{-2}\mathrm{ml}$

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JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

millimole of $\mathrm{NaOH}=0.24\times 25$

For nuetralisation, equal moles of monoacidic base and monobasic acid must be present:

$\therefore$ millimole of acid $=0.24\times 25$

$\Rightarrow$ mass of acid $=0.24\times 25\times 24.2g$

for pure acid,

$V=\frac{w}{d};(d=1.21\mathrm{kg}/L=1.21g/\mathrm{ml})$

$\therefore V=\frac{0.24\times 25\times 24.2}{1.12}\times {10}^{-3}$

$=120\times {10}^{-3}\mathrm{ml}$

$=12\times {10}^{-2}\mathrm{ml}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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JEE Main 2023 — 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