Redox and Equivalent ConceptMedium

Question

x g of KHC2O4 requires 100 ml of 0.02 M-KMnO4 in acidic medium. In another experiment, y g of KHC2O4 requires 100 ml of 0.05 M-Ca(OH)2. The ratio of x and y is

Options

A.1: 1
B.1: 2
C.2: 1
D.5: 4

Solution

$n_{eq}MnO_{4}^{-} = n_{eq}FeSO_{4}$

Or $\frac{x}{M} \times 2 = \frac{100 \times 0.02}{1000} \times 5(1)$

$n_{eq}KHC_{2}O_{4} = n_{eq}Ca(OH)_{2} $$${\text{Or }\frac{y}{M} \times 1 = \frac{100 \times 0.5}{1000} \times 2(2) }{\therefore\frac{x}{y} = \frac{1}{2}}$$

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