Question

What is the equivalent mass of $A{s}_2{S}_3$ in the following reactions? (Molar mass of $A{s}_2{S}_3=M$)
$A{s}_2{S}_3+N{O}_3^{-}+H^{+}\to A{s}_2{O}_5+N{O}_2^{-}+S{O}_3+H_{2}0$

• A. $M/4$
• B. $M/24$
• C. $M/28$
• D. $M/12$

Answer 1

The correct option is C $M/28$

$N_f=\sum \left(\Delta OS\right)$

$N{O}_3{}^{-}+A{s}_2{S}_3⟶As{O}_4^{-3}+NO+S{O}_2^{-4}$

Now,$N_f=\left(2x2\right){}_{forAS}+\left(8x3\right){}_{forS}=28$

$\therefore$ Equivalent weight of $A{s}_2{S}_3=\frac{M}{28}$