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

What volume of 0.02 M acidified KMnO4 is required to completely oxidise 1.44 g of ferrous oxalate, FeC2O4 (molar mass 143.9 g mol^-1)?

Single correct option

What volume of

0.02\ M

acidified

\mathrm{KMnO_4}

is required to completely oxidise

1.44\ \mathrm{g}

of ferrous oxalate,

\mathrm{FeC_2O_4}

(molar mass

143.9\ \mathrm{g\,mol^{-1}}

Show answer and solution

Answer: (C) $300\ \mathrm{mL}$

Moles of

\mathrm{FeC_2O_4}=1.44/143.9=0.0100

. Both ions are oxidised:

\mathrm{Fe^{2+}\to Fe^{3+}}

e^-

) and

\mathrm{C_2O_4^{2-}\to 2CO_2}

e^-

), so

n

-factor

=3

and equivalents

=0.030

\mathrm{MnO_4^-\to Mn^{2+}}

takes

5

electrons, so moles of

\mathrm{KMnO_4}=0.030/5=0.0060

, and

V=0.0060/0.02=0.30

=300

mL. Counting only the oxalate (

n=2

) would give

200

mL.

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake