KCET · Chemistry · Redox Reactions
The volume of \(0.1 \mathrm{M}\) oxalic acid that can be completely oxidized by \(20 \mathrm{~mL}\) of \(0.025 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) solution is
- A \(25 \mathrm{~mL}\)
- B \(12.5 \mathrm{~mL}\)
- C \(37.5 \mathrm{~mL}\)
- D \(125 \mathrm{~mL}\)
Answer & Solution
Correct Answer
(B) \(12.5 \mathrm{~mL}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}=\frac{\text { molecular weight }}{\text { equivalent weight }} \times \mathrm{M}\)
\[
\begin{aligned}
\mathrm{N}_{\mathrm{KMnO}_{4}} &=\frac{158}{31.6} \times 0.025 \\
&=0.125 \\
\mathrm{~N}_{\mathrm{O} . \mathrm{A}} &=\frac{126}{63} \times 0.1=0.2 \\
\because \quad \mathrm{N}_{\mathrm{V}} \mathrm{V}_{1} &=\mathrm{N}_{2} \mathrm{~V}_{2} \\
0.2 \times \mathrm{V} &=0.125 \times 20 \\
\therefore \quad & 0.125 \times 20 \\
\mathrm{~V} &=0.2 \\
&=12.5 \mathrm{~mL}
\end{aligned}
\]
\[
\begin{aligned}
\mathrm{N}_{\mathrm{KMnO}_{4}} &=\frac{158}{31.6} \times 0.025 \\
&=0.125 \\
\mathrm{~N}_{\mathrm{O} . \mathrm{A}} &=\frac{126}{63} \times 0.1=0.2 \\
\because \quad \mathrm{N}_{\mathrm{V}} \mathrm{V}_{1} &=\mathrm{N}_{2} \mathrm{~V}_{2} \\
0.2 \times \mathrm{V} &=0.125 \times 20 \\
\therefore \quad & 0.125 \times 20 \\
\mathrm{~V} &=0.2 \\
&=12.5 \mathrm{~mL}
\end{aligned}
\]
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