Question

${As}_2{O}_3$ is oxidised to $H_{3}As{O}_4$ by $KMn{O}_4$ in acidic medium. Volume of $0.02M$ $KMn{O}_4$ required for this purpose by $1$ mmol of ${As}_2{O}_3$ will be:

• A. $10mL$
• B. $25mL$
• C. $40mL$
• D. $80mL$

Answer 1

Answer: (B)

$25mL$

$A{s}_2{O}_3\to H_{3}As{O}_4+4e^{-}$

$Mn{O}_4^{-}+5e^{-}\to M{n}^{2+}$

'n' factor for $A{s}_2{O}_3=4$, while that for $KMn{O}_4=5$

meq of $KMn{O}_4=$meq of $A{s}_2{O}_3$

$5\times molarity\times volume=moles\times n-factor$

$\Rightarrow 5\times 0.02\times V=1\times 4\times {10}^{-3}$

$\Rightarrow V=25ml$