Question

10 mL of H₂O₂ solution on treatment with KI and titration of liberated I₂, required 10 mL of 1N hypo. Thus H₂O₂ is –

• A. 1 N
• B. 5.6 volume
• C. 17 g L^–1
• D. All are correct

Answer 1

Answer: (D)

All are correct

Answer 2

H₂O₂ + 2I ^– ® I₂

I₂ + 2S₂O${}_{3}^{2–}$® S₄O${}_{6}^{2–}$+ 2I^–

H₂O₂ º I₂ º 2S₂O${}_{3}^{2–}$

N₁V₁ = N₂V₂

H₂O₂ hypo

N₁(H₂O₂) = $\frac{10 \times 1}{10}$= 1N

Conc. = N × E = 17 g/litre

Volume strength = 5.6 × normality

= 5.6 × 1

= 5.6 volume