A 2.0 g sample containing $\ce{MnO2}$ is treated with HCl liberating $\ce{Cl2}$. The $\ce{Cl2}$ gas is passed into a solution of KI and 60.0 mL of 0.1 M $\ce{Na2S2O3}$ is required to titrate the liberated iodine. The percentage of $\ce{MnO2}$ in the sample is ______. (Nearest integer) [Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Redox Reactions and Electrochemistry

A 2.0 g sample containing

MnO X_{2}

is treated with HCl liberating

Cl X_{2}

. The

Cl X_{2}

gas is passed into a solution of KI and 60.0 mL of 0.1 M

Na X_{2} S X_{2} O X_{3}

is required to titrate the liberated iodine. The percentage of

MnO X_{2}

in the sample is ______.
(Nearest integer)
[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

Answer: 13

The key is to work backwards from the thiosulphate titration:

Millimoles of

Na X_{2} S X_{2} O X_{3}

60.0 \times 0.1 = 6

mmol

From

I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}

: mmol of

I X_{2}

6/2 = 3

mmol

From

Cl X_{2} + 2 I X^{-} 2 Cl X^{-} + I X_{2}

: mmol of

Cl X_{2}

= 3 mmol

From

MnO X_{2} + 4 HCl MnCl X_{2} + Cl X_{2} + 2 H X_{2} O

: mmol of

MnO X_{2}

= 3 mmol

Mass of

MnO X_{2}

3 \times 1 0^{- 3} \times 87 = 0.261

g

Percentage =

(0.261/2.0) \times 100 = 13.05% \approx 13%

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer