When equimolar mixture of $C u_{2} S$ and $C u S$ reacts with $B a {(M n O_{4})}_{2}$ in acidic medium the final products formed are $C u^{2 +}, S O_{2}, M n^{2 +}$ . If molecular weights of $C u_{2} S, C u S$ and $B a {(M n O_{4})}_{2}$ are $M_{1}, M_{2}$ and $M_{3}$ respectively. Then $E_{W} = E q u a l e n t w e i g h t$

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$C u_{2} S a n d C u S h a v e s a m e E w' s$

$E w o f C u S i s \frac{M_{2}}{6}$

$E w o f B a (M n O_{4})_{2} i s \frac{M_{3}}{5}$

$E w o f C u S i s \frac{M_{1}}{8}$

answer is A, B.

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Detailed Solution

$\begin{array}{l} C u_{2}^{+ 1} S^{- 2} \to 2 C u^{2 +} + S^{+ 4} O_{2} n_{+} = 8 ∴ \frac{M_{1}}{8} \\ C u^{+ 2} S^{- 2} \to C u^{2 +} + S^{+ 4} O_{2} n_{+} = 6 ∴ \frac{M_{2}}{6} \\ B^{+ 2} a {(M n^{+ 7} O_{4})}_{2} \to B a^{2 +} + 2 M n^{+ 2} n_{+} = 10 \frac{M_{3}}{10} \end{array}$