To a 25 ml H₂O₂ solution, excess of acidified solution of KI was added. The iodine liberated required 20 ml of 0.3 N Na₂S₂O₃ solution. The volume strength of H₂O₂ solution is:

2KI + H₂SO₄ + H₂O₂ → K₂SO₄ + 2H₂O + I₂
2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI
milli eq. of H₂O₂ in 25ml = milli eq. of I₂ = milli eq. of Na₂S₂O₃
milli eq. of H₂O₂ in 25ml = 20 × 0.3 = 6
milli eq. of H₂O₂ in 1000ml = (6/25) × 1000 = 240
Equivalent per litre = 240/1000
Gram per litre of H₂O₂ = (240/1000) × 17 = 4.08g/L (Equivalent weight of H₂O₂ = 34/2 = 17)
Hence, option D is correct.