10.0 mL of a commercial $H X_{2} O X_{2}$ solution is diluted to 250 mL. A 25.0 mL portion of the diluted solution requires 18.0 mL of 0.0200 M $KMnO X_{4}$ in acidic medium. Find the volume strength of the original solution, i.e. litres of $O X_{2}$ at STP from 1 L of solution by $2 H X_{2} O X_{2} 2 H X_{2} O + O X_{2}$ (molar volume 22.4 L/mol). Give the answer to two decimal places.

Numerical answer type, answer in volumes, tolerance 0.05.

Answer: 10.08 volumes

mmol

MnO X_{4} X^{-}

18.0 \times 0.0200 = 0.360

;

H X_{2} O X_{2}

in 25.0 mL =

0.360 \times 2.5 = 0.900

mmol, so 9.00 mmol in 250 mL, all from 10.0 mL of original. Molarity = 9.00/10.0 = 0.90 M. Each mole gives 0.5 mol

O X_{2}

= 11.2 L, so volume strength = 0.90 x 11.2 = 10.08.