Question

Potassium acid oxalate $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ can be oxidized by $Mn{O}_4^{-}$ in acidic medium. Calculate the volume of $0.3MMn{O}_4^{-}$ reacting in acid solution with $10.16g$ of the acid oxalate.
(Molar mass of potassium acid oxalate is $508gmo{l}^{-1}$.)

• A. $106.7mL$
• B. $160.0mL$
• C. $1.06L$
• D. $0.16L$

Answer 1

The correct option is A $106.7mL$

$\underset{\text{Potassium acid oxalate}}{K_2{\stackrel{+3}{C}}_2{O}_4.3H_2{\stackrel{+3}{C}}_2{O}_4.4H_{2}O}+\stackrel{+7}{M}n{O}_4^{-}\stackrel{H^{+}}{⟶}\stackrel{+2}{M}{n}^{2+}+\stackrel{+4}{C}{O}_2$

$n_f=\left(|\text{Change in O.S.}|\right)\times \text{Number of atoms}$
$n_{f_{\text{potassium acid oxalate}}}=(|+3-(+4\left)|\right)\times 8=8$
$n_{f_{Mn{O}_4^{-}}}=(|+7-(+2\left)|\right)\times 1=5$

From the law of equivalence,
$\text{Equivalents of potassium acid oxalate}=\text{Equivalents of}Mn{O}_4^{-}$
$\frac{W_{\text{potassium acid oxalate}}}{\text{Molar mass of potassium acid oxalate}}\times n_{f_{\text{potassium acid oxalate}}}=M_{Mn{O}_4^{-}}\times V_{Mn{O}_4^{-}}\times n_{f_{Mn{O}_4^{-}}}$
$\frac{10.16g}{508gmo{l}^{-1}}\times 8=\left(0.3mol{L}^{-1}\right)\times V_{Mn{O}_4^{-}}\times 5$
$\therefore V_{Mn{O}_4^{-}}\approx 0.1067L\approx 106.7mL$