Question

In the following reaction,
$N{O}_3^{⊝}+A{s}_2{S}_3+H_{2}O\to As{O}_4^{3-}+NO+S{O}_4^{2-}+H^{⨁}$, the equivalent weight of $A{s}_2{S}_3$ (with molecular weight $M$) is:

• A. $\frac{3M}{28}$
• B. $\frac{M}{4}$
• C. $\frac{M}{24}$
• D. $\frac{M}{28}$

Answer 1

The correct option is D $\frac{M}{28}$

$N{O}_3^{⊝}+A{s}_2{S}_3\to As{O}_4^{3-}+NO+S{O}_4^{2-}$
$A{s}_2{S}_3+20H_{2}O\to 2As{O}_4^{3-}+3S{O}_4^{2-}+40H^{⨁}+28e^{-}$
$E_{A{s}_2{S}_3}=\frac{M}{28}$ ($n$ factor $=28$) (electrons lost/mole)