Question

$0.3\, g$ of an oxalate salt was dissolved in $100\, mL$ solution. The solution required $90\, mL$ of $N / 20 KMnO _{4}$ for complete oxidation. The $\%$ of oxalate ion in salt is

• A. $33 \%$
• B. $66 \%$
• C. $70 \%$
• D. $40 \%$

Answer 1

Answer: (B)

Meq. of $KMnO _{4}= M$ eq. of $C _{2} O _{4}^{-2}$

$90 \times \frac{1}{20}=100 \times N_{ C _{2} O _{4}^{-2}}$

$=\frac{9}{2 \times 2}=\frac{9}{4}$

Wt of oxalate $=\frac{9}{4} \times 88 \times 10^{-3}=22 \times 9 \times 10^{-3}=198 \times 10^{-3}$

$\% C _{2} O _{4}^{-2}=\frac{.198}{.300} \times 100=66 \%$