Question

28 ml of 0.1M oxalic acid (H₂C₂O₄) solution requires 10 ml of KMnO₄ for titration. 10 ml of this sample of KMnO₄ when added to excess of NH₂OH (hyroxyl amine) liberates N₂ at STP. The volume of N₂ liberated at NTP is –

• A. 24 ml
• B. 38 ml
• C. 46 ml
• D. 56 ml

Answer 1

Answer: (B)

38 ml

Answer 2

In acidic medium MnO₄^– ® Mn⁺²

So $(NV)_{KMnO_{4}}$= $(NV)_{H_{2}C_{2}O_{4}}$

N × 10 = 28 × 0.1 × 2

N = 0.56

molarity KMnO₄ in acidic medium = $\frac{0.56}{5}$

with NH₂OH (medium is weakly basic).

$\overset{–1}{NH_{2}OH} \rightarrow \overset{0}{N_{2}}(NV)_{KMnO_{4}}$ = $\frac{w}{E} \times 1000$

$\left( \frac{0.56}{5} \times 3 \right) \times 10$ = $\frac{w}{M/2}$ × 1000

mole of N₂ = 1.68 × 10^–3

volume at STP = 1.68 × 10^–3 × 22400

= 37.63 ${}_{–}^{\sim}$ 38 ml.