In the following reaction (unbalanced), equivalent weight of $A s_{2} S_{3}$ is related to molecular weight $M$ by M/x.
${A s}_{2} S_{3} + H^{+} + {N O}_{3}^{-} ⟶ N O + H_{2} O + A s O_{4}^{3 -} + S O_{4}^{2 -}$ .
value of x is _______.

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Solution

Reaction:
${A s}_{2} S_{3} 2 A s = + 3 * 2 = + 6 3 S = - 2 * 3 ⟶ A s O_{4}^{3 -} 2 \times 5 = 10 + 3 S O_{4}^{3 -} 3 \times 6 = 18$
net change in oxidation of $A s_{2} S_{3} = 18 + 4 + 6 = 28$
Thus, equivalent mass of $A_{2} S_{3} = \frac{Molar mass}{change in oxidation} = \frac{M}{28}$

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