JEE Main 2025 · Chemistry · Mole Concept and Stoichiometry · medium

0 mL of a commercial H2O2 sample is diluted to 100 mL. 10 mL of the diluted solution exactly decolourises 20 mL of 0.1 M acidified KMnO4. The…

Numerical answer

5.0\ \mathrm{mL}

of a commercial

\mathrm{H_2O_2}

sample is diluted to

100\ \mathrm{mL}

10\ \mathrm{mL}

of the diluted solution exactly decolourises

20\ \mathrm{mL}

0.1\ \mathrm{M}

acidified

\mathrm{KMnO_4}

. The concentration of

\mathrm{H_2O_2}

in the commercial sample, in

\mathrm{g\,L^{-1}}

, is (molar mass of

\mathrm{H_2O_2}=34

)

Show answer and solution

Answer: 340

\mathrm{2MnO_4^-+5H_2O_2+6H^+\to2Mn^{2+}+5O_2+8H_2O}

. Moles of

\mathrm{KMnO_4}=0.020\times0.1=2\times10^{-3}

, so moles of

\mathrm{H_2O_2}=2.5\times2\times10^{-3}=5\times10^{-3}

10

mL, i.e.

0.5\ \mathrm{M}

in the diluted solution. Dilution factor

100/5=20

, so the commercial sample is

10\ \mathrm{M}

. Mass concentration

=10\times34=340\ \mathrm{g\,L^{-1}}

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