In the reaction of KMnO_ 4 with an oxalate in acidic medium, MnO_ 4 ^ - is reduced to Mn^ 2+ and C_ 2 O_ 4 ^ 2- is oxidised to CO_ 2 . Hence, 50 mL of 0.02 M KMnO_ 4 is equivalent to

Correct answer: B. 50 mL of 0.05 M H_ 2 C_ 2 O_ 4 eg. of (KMnO_ 4 ) aligned &=5 ×0.02 ×50 &=250 ×0.02 &=509. aligned ) eq. of (KAnO_ 4 eq_ 1 ) of (C_ 2 O_ 4 _ 2 -) aligned 5=option 2 &=50 ×0.05 ×2 &=589. aligned

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

BITSAT 2012 · Chemistry · Some Basic Concepts of Chemistry

In the reaction of KMnO_ 4 with an oxalate in acidic medium, MnO_ 4...

In the reaction of

\mathrm{KMnO}_{4}

with an oxalate in acidic medium,

\mathrm{MnO}_{4}^{-}

is reduced to

\mathrm{Mn}^{2+}

and

\mathrm{C}_{2} \mathrm{O}_{4}^{2-}

is oxidised to

\mathrm{CO}_{2}

. Hence,

50 \mathrm{ mL}

of 0.02

\mathrm{M} \mathrm{KMnO}_{4}

is equivalent to In the reaction of KMnO_ 4 with an oxalate in acidic medium, MnO_ 4 ^ - is reduced to Mn^ 2+ and C_ 2 O_ 4 ^ 2- is

Options

A $100 \mathrm{ mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
B $50 \mathrm{ mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
C $25 \mathrm{ mL}$ of $0.2 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
D $50 \mathrm{ mL}$ of $0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$

Correct answer

B. $50 \mathrm{ mL}$ of $0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$

Step-by-step solution

eg. of (\mathrm{KMnO}_{4})

\begin{aligned}&=5 \times0.02 \times50 \\ &=250 \times0.02 \\ &=509. \end{aligned}

)

eq.

of

(\mathrm{KAnO}_{4} \equiv \mathrm{eq}_{1}) of (\mathrm{C}_{2} \mathrm{O}_{4}{ }_{2}-) In the reaction of KMnO_ 4 with an oxalate in acidic medium, MnO_ 4 ^ - is reduced to Mn^ 2+ and C_ 2 O_ 4 ^ 2- is

\begin{aligned}5=\operatorname{option} 2 &=50 \times0.05 \times2 \\ &=589. \end{aligned}

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