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Q.

$K M n O_{4}$ Reaction with oxalic acid according to the equation
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ here, 20 mL of 0.1 M is equivalent to

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a

120mL of 0.25 $M H_{2} C_{2} O_{4}$

b

150mL of 0.10 $M H_{2} C_{2} O_{4}$

c

25mL of 0.20 $M H_{2} C_{2} O_{4}$

d

50mL of 0.20 $M H_{2} C_{2} O_{4}$

answer is C.

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Detailed Solution

N₁V₁ = N₂V₂ = 0.5 × 20 = 0.4 × 25.

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