100 mL of M/10 $B a {(M n O_{4})}_{2}$ in acidic medium can be oxidised completely with

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100mL of 1M $F e S O_{4}$ solution

$\frac{100}{3} m L$ of 1M $F e C_{2} O_{4}$ solution

25mL of 1M $K_{2} C r_{2} O_{7}$ solution

75mL of 1M $C_{2} O_{4}^{2 -}$ solution

answer is A, B.

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Detailed Solution

$\begin{array}{l} (10 e^{-} + 2 {MnO}_{4}^{-} \to 2 {Mn}^{2 +}) (n = 10) \\ a) {Fe}^{2 +} \to {Fe}^{3 +} + e^{-} (n = 1) \\ mEq of Ca {({MnO}_{4})}_{2} = 100 \times \frac{1}{10} \times 10 = 100 \\ mEq of {FeSO}_{4} = 100 \times 1 = 100 \end{array} \begin{array}{l} b) mEq of {FeC}_{2} O_{4} (n = 3) \\ [\begin{array}{l} {Fe}^{+ 2} \to {Fe}^{+ 3} + e^{-} \\ C_{2} O_{4}^{2 -} \to 2 {CO}_{2} + 2 e^{-} \end{array}] \\ = \frac{100}{3} \times 3 \times 1 \\ = 100 = mEq of Ba {({MnO}_{4})}_{2} \end{array} \begin{array}{l} c) mEq of K_{2} {Cr}_{2} O_{7} (n = 6) = 25 \times 1 \times 6 = 150 \\ d) mEq of C_{2} O_{4}^{2 -} (n = 2) = 75 \times 2 \times 1 = 150 \end{array}$