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Q.

$40 mL 0.1 N {KMnO}_{4}$ is equivalent to $30 mL {KHC}_{2} O_{4}$ solution. How many mL of $0.1 N KOH$ are required to titrate 60mL of same ${KHC}_{2} O_{4}$ solution?

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a

40 mL

b

35.5 ML

c

28.57 mL

d

30 mL

answer is A.

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Detailed Solution

$\begin{array}{l} 40 \times 0.1 = 30 \times M \times 2 \\ M = \frac{4}{15 \times 4} = \frac{1}{15} \\ V = 40 ml \end{array}$

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