When a equimolar mixture of $C u_{2} S$ and CuS is titrated with $B a {(M n O_{4})}_{2}$ in acidic medium, the final products contain $C u^{2 +}, S O_{2} a n d M n^{2 +} .$ If the mol. mass of $C u_{2} S, C u S a n d B a {(M n O_{4})}_{2}$ are $M_{1,} M_{2} a n d M_{3}$ respectively then:

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eq. mass of $C u_{2} S$ is $\frac{M_{1}}{8}$

eq mass of CuS is $\frac{M_{2}}{6}$

eq. mass of $B a {(M n O_{4})}_{2}$ is $\frac{M_{3}}{5}$

$C u_{2} S a n d C u S$ both have same equivalents in mixture

answer is A, B.

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Detailed Solution

Let $n_{C u_{2} S} = a; n_{C u S} = a$

$\begin{matrix} C u_{2} S \\ n - f a c t o r = 8 \end{matrix} + \begin{matrix} C u S \\ 6 \end{matrix} + \begin{matrix} B a {(M n O_{4})}_{2} \\ 10 \end{matrix} + H^{+} \to C u^{2 +} + S O_{2} + M n^{2 +}$

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