Question

X gram of pure $A{s}_2{S}_3$ is completely oxidised to respective highest oxidation states by $50$ ml of $0.1$ M hot acidified $KMn{O}_4$ then mass of $A{s}_2{S}_3$ (X) taken is:

(Molar mass of $A{s}_2{S}_3=246$)

• A. 22.4 g
• B. 0.22 g
• C. 64.23 g
• D. None

Answer 1

Answer: (B)

0.22 g

$5A{s}_2{S}_3+28KMn{O}_4+H^{+}\to 10H_{3}As{O}_4+28M{n}^{2+}+S{O}_4^{2-}$

M.eq of $A{s}_2{S}_3$ = M.eq of $KMn{O}_4$

$\frac{x}{246}\times 1000\times 28=50\times 0.1\times 5$

After solving we get,

x = 0.22 g

Hence, option B is correct .