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Q.

If M represents molecular mass of ${Mn}_{3} O_{4}$ hen what will be its equivalent mass if it undergoes disproportionation reaction as shown:
${Mn}_{3} O_{4} ⟶ {MnO}_{4}^{-} + {Mn}^{2 +}$

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a

$\frac{M}{13}$

b

$\frac{15 M}{26}$

c

$\frac{M}{2}$

d

$\frac{26 M}{15}$

answer is C.

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Detailed Solution

$M n_{3} O_{4} ⟶ M n O_{4}^{-} + M n^{2 +}$

$M n_{3} O_{4} ⟶ 3 M n O_{4}^{-} + 13 e^{-}$

$2 e^{-} + M n_{3} O_{4} ⟶ 3 M n^{2 +}$

$n_{eff} = \frac{2 \times 13}{2 + 13} = \frac{26}{15}$

$E = \frac{M}{26} \times 15$ = $\frac{15 M}{26}$

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