Question

A 0.1097 g sample of $A{s}_2{O}_3$ required 26.10 mL of acidic $KMn{O}_4$ solution for its titration. The molarity of $KMn{O}_4$ solution is:

• A. $0.02$
• B. $0.04$
• C. $0.018$
• D. $0.3$

Answer 1

Answer: (C)

$0.018$

$5A{s}_2{O}_3+4Mn{O}_4^{-}+9H_{2}O+12H^{+}\to 10H_{3}As{O}_4+4M{n}^{(2+)}$
5 moles $A{s}_2{O}_3$ reacts with 4 moles of $KMn{O}_4$.
Molar mass of $A{s}_2{O}_3$: 197.841 g/mol
0.1097g $A{s}_2{O}_3=$ 0.000554 mole $A{s}_2{O}_3$.
Thus the molarity of the $KMn{O}_4\left(aq\right)$ is $=\frac{0.000554\times 4}{5\times 0.0261}=0.01698M$, closest to which is option (C).