$28 {N O_{3}}^{-} + 3 A s_{2} S_{3} + 4 H_{2} O \to 6 {A s O_{4}}^{3 -} + 28 N O + 9 {S O_{4}}^{2 -} + 8 H^{+}$
What will be the equivalent mass of $A s_{2} S_{3}$ in above reaction?

\frac{M . w t .}{2}

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\frac{M . w t .}{4}

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\frac{M . w t .}{24}

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\frac{M . w t .}{28}

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Solution

The correct option is A $\frac{M . w t .}{28}$
In the reaction,
$A s_{2} S_{3} \to A s O_{4}^{3 -}$
no. of electron change = 2
so, n-factor = no. of electron change * no. of atom = 6 * 2 = 12

equivalent mass = $\frac{m o l e c u l a r m a s s}{n - f a c t o r}$ = $\frac{m o l e c u l a r m a s s}{12}$

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Similar questions

28 {N O_{3}}^{-} + 3 A s_{2} S_{3} + 4 H_{2} O \to 6 {A s O_{4}}^{3 -} + 28 N O + 9 {S O_{4}}^{2 -} + 8 H^{+}

A s_{2} S_{3}

{28 N O}_{3}^{-} + {3 A s}_{2} S_{3} + {4 H}_{2} O ⟶ {6 A s O}_{4}^{3 -} + 28 N O + {9 S O}_{4}^{2 -} + H^{+}

{A s}_{2} S_{3}

28 {N O}_{3}^{-} + 3 A s_{2} S_{3} + 4 H_{2} O \to 6 A s O_{3}^{3 -} + 28 N O + 9 {S O}_{4}^{2 -} + 8 H^{+}

{A s}_{2} S_{3}

? (

{A s}_{2} S_{3} = M)

(M . w t = 56)

K_{s p}

C a S O_{4} \cdot 5 H_{2} O

9 \times 10^{- 6}

1 g

C a S O_{4} (M . W t . = 136)

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