$10.0 mL$ of hydrogen peroxide is diluted to $100 mL$ . A $20.0 mL$ portion is treated with excess $KI$ and acid ( $H X_{2} O X_{2} + 2 I X^{-} + 2 H X^{+} I X_{2} + 2 H X_{2} O$ ), and the iodine needs $15.2 mL$ of $0.100 M$ thiosulphate. Taking 1 mol of $H X_{2} O X_{2}$ to give $11.2$ L of $O X_{2}$ at STP on decomposition, the volume strength of the original solution is

A8.51 V
B4.26 V
C0.85 V
D2.13 V

Answer: B

I2 in 20.0 mL = (0.0152 x 0.100)/2 = 7.60e-04 mol = mol H2O2. In 100 mL: 3.80e-03 mol, which was in 10.0 mL, so M = 0.380. Volume strength = M x 11.2 = 4.26 V.

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