Question

Potassium acid oxalate $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ having molar mass 508 g/mol can be oxidized by $Mn{O}_4^{-}$ in acidic medium. Calculate the volume of 0.1 M $KMn{O}_4$ reacting in acid solution with one gram of the acid oxalate.

• A. 52.80 mL
• B. 15.84 mL
• C. 7.92 mL
• D. 31.68 mL

Answer 1

The correct option is D 31.68 mL

$K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O;nf=8$ as each of the 8 carbon atoms in the molecule undergo a +1 change in oxidation state.
$nf$ of permangante ion is 5 as manganese gets reduced from +7 oxidation state to +2 oxidation state.
$\text{Equivalents of oxalate = Equivalents of}KMn{O}_4$
$\frac{1}{\text{molecular weight}}\times 8=V\times 0.1\times 5$
Molecular mass of Potassium acid oxalate = 508 g/mol
$\Rightarrow V=\frac{8}{508\times 0.1\times 5}=0.03168L$
$\Rightarrow V=31.68ml$