JEE Main 2018 · Chemistry · Redox Reactions and Titrations · medium

10 mL of a hydrogen peroxide sample is diluted to 100 mL. 20 mL of the diluted solution decolourises 25 mL of 0.04 M acidified KMnO4. The volume…

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10\ \mathrm{mL}

of a hydrogen peroxide sample is diluted to

100\ \mathrm{mL}

20\ \mathrm{mL}

of the diluted solution decolourises

25\ \mathrm{mL}

0.04\ M

acidified

\mathrm{KMnO_4}

. The volume strength of the original sample (volume of

\mathrm{O_2}

at STP from

1\ \mathrm{mL}

\mathrm{H_2O_2}

on decomposition) is

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Answer: (C) $14$

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

. Moles

\mathrm{KMnO_4}=25\times0.04=1.0

mmol, so

\mathrm{H_2O_2}=\tfrac52\times1.0=2.5

mmol in

20

mL of the diluted solution:

0.125\ M

. Undoing the ten-fold dilution gives

1.25\ M

for the original. Since

2\mathrm{H_2O_2\to2H_2O+O_2}

1\ M

gives

11.2

mL of

\mathrm{O_2}

(STP) per mL, i.e.

11.2

volume. Strength

=1.25\times11.2=14

volume. Forgetting the dilution gives

1.4

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