KMn O_4 reacts with oxalic acid according to the equation, 2MnO_4… - Vidyadip

MCQ

$KMn{O_4}$ reacts with oxalic acid according to the equation, $2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$, here $20\, ml$ of $ 0.1\, M$ $KMn{O_4}$ is equivalent to

A
$20\, ml $ of $0.5\, M$ ${H_2}{C_2}{O_4}$
✓
$50\, ml$ of $0.1\, M$ ${H_2}{C_2}{O_4}$
C
$50\, ml$ of $0.5\, M$ ${H_2}{C_2}{O_4}$
D
$20\, ml$ of $0.1\, M$ ${H_2}{C_2}{O_4}$

✓

Answer

Correct option: B.

$50\, ml$ of $0.1\, M$ ${H_2}{C_2}{O_4}$

b
(b) $KMn{O_4}$ Oxalic acid

$\frac{{{M_1}{V_1}}}{{{n_1}}} = $$\frac{{{M_2}{V_2}}}{{{n_2}}}$; $\frac{{20 \times 0.1}}{2} = \frac{{{M_2}{V_2}}}{5}$; ${M_2}{V_2} = 5$.

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