Question

Pottasium acid oxalate, $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ can be oxidized by $KMn{O}_4$ in acid medium. Calculate the volume (in mL) of $0.1$ $MKMn{O}_4$ that reacts with one gram of the acid oxalate in acidic medium.

• A. $20$
• B. $40$
• C. $23$
• D. None of the above

Answer 1

The correct option is D None of the above

One mole of potassium acid oxalate contains $4$ moles of oxalate ions.
Two moles of permanganate ions oxidize five moles of oxalate ions.

$4$ moles of oxalate ions will be oxidized by $\frac{2}{5}\times 4=1.6$ moles of permagnate ions

The molar mass of potassium acid oxalate is $328$ g/mol

One gram of potassium acid oxalate corresponds to $\frac{1}{328}=0.0030487$ moles

$0.0030487$ moles of potassium acid oxalate will be oxidised by $0.0030487\times 1.6=0.004878$ moles of oxalate ions.

Volume of $0.1MKMn{O}_4$ solution required $=\frac{0.004878moles}{0.1moles/L}=0.04878L=48.8mL$