JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Redox Reactions and Electrochemistry

To measure the quantity of

MnCl_{2}

dissolved in an aqueous solution, it was completely converted to

KMnO_{4}

using the reaction,

MnCl_{2} + K_{2} S_{2} O_{8} + H_{2} O \to KMnO_{4} + H_{2} SO_{4} + HCl

(equation not balanced).

Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of

MnCl_{2}

(in mg) present in the initial solution is ____.

(Atomic weights in

g mol^{- 1}

Mn = 55

Cl = 35.5

)

Answer: 126

Solution

The number of milliequivalents of

MnCl_{2}

equals the number of milliequivalents of

KMnO_{4}

, which equals the number of milliequivalents of oxalic acid. Oxalic acid (

H_{2} C_{2} O_{4}

, MW = 90, n-factor = 2) in 225 mg: meq =

\frac{225}{90} \times 2 = 5

meq. For

MnCl_{2}

(MW = 126, n-factor = 5 for conversion to

MnO_{4}^{-}

): weight =

5 \times 1 0^{- 3} \times \frac{126}{5} = 126

mg.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Redox Reactions and Electrochemistry

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