Question

$Mn{O}_4^{-}$ reacts with oxalate according to the equation:
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
Here $20mL$ of $0.1MMn{O}_4^{-}$ is equivalent to -

• A. $20mL$ of $0.5M{C}_2{O}_4^{2-}$
• B. $50mL$ of $0.5M{C}_2{O}_4^{2-}$
• C. $50mL$ of $0.1M{C}_2{O}_4^{2-}$
• D. $20mL$ of $0.1M{C}_2{O}_4^{2-}$

Answer 1

The correct option is C $50mL$ of $0.1M{C}_2{O}_4^{2-}$

$2\stackrel{+7}{M}n{O}_4^{-}+5{\stackrel{+3}{C}}_2{O}_4^{2-}+16H^{+}\to 2\stackrel{+2}{M}{n}^{2+}+10\stackrel{+4}{C}{O}_2+8H_{2}O$

$n_f=\left(|\text{Change in O.S.}|\right)\times \text{Number of atoms}$
$n_{f_{Mn{O}_4^{-}}}=|+2-(+7)|\times 1=5$
$n_{f_{C_2{O}_4^{2-}}}=|+4-(+3)|\times 2=2$

From the law of equivalence,
$\text{Milliequivalents of}Mn{O}_4^{-}=\text{Milliquivalents of}{C}_2{O}_4^{2-}$
$0.1\times 20\times 5=M_{C_2{O}_4^{2-}}\times V_{C_2{O}_4^{2-}}\left(\text{in}mL\right)\times 2$
$\therefore M_{C_2{O}_4^{2-}}\times V_{C_2{O}_4^{2-}}\left(\text{in}mL\right)=5$
Hence option (c) is correct.