JEE Main 2016 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

Taking 20-volume hydrogen peroxide to mean that 1 mL liberates 20 mL of O2 at STP (molar volume 22.4 L) on decomposition, the volume of 0.10 M…

Single correct option

Taking

20

-volume hydrogen peroxide to mean that

1\ \mathrm{mL}

liberates

20\ \mathrm{mL}

\mathrm{O_2}

at STP (molar volume

22.4\ \mathrm{L}

) on decomposition, the volume of

0.10\ \mathrm{M}

acidified

\mathrm{KMnO_4}

needed to oxidise

10.0\ \mathrm{mL}

of this

\mathrm{H_2O_2}

solution is nearly

Show answer and solution

Answer: (B) $71\ \mathrm{mL}$

2\mathrm{H_2O_2}\to2\mathrm{H_2O}+\mathrm{O_2}

1

mL gives

20

\mathrm{O_2}

=20/22400

mol, so moles of

\mathrm{H_2O_2}

per mL

=2\times20/22400=1.786\times10^{-3}

, molarity

=1.786

M. In

10.0

mL:

0.01786

mol. In acid,

2\mathrm{MnO_4^-}+5\mathrm{H_2O_2}+6\mathrm{H^+}\to2\mathrm{Mn^{2+}}+5\mathrm{O_2}+8\mathrm{H_2O}

, so moles

\mathrm{MnO_4^-}=0.01786\times2/5=7.14\times10^{-3}

. Volume

=7.14\times10^{-3}/0.10=71.4

mL. A 1:1 ratio gives

179

mL and an inverted ratio gives

446

mL.

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