10.0 mL of a H2O2 solution is added to excess acidified KI (H2O2 + 2I- + 2H+ -> I2 + 2H2O). The iodine liberated needs 22.0 mL of 0.100 M Na2S2O3. The molarity of H2O2 is:

JEE Main levelSingle correct optionDifficulty 2 of 3eq139

10.0 mL of a $H X_{2} O X_{2}$ solution is added to excess acidified KI ( $H X_{2} O X_{2} + 2 I X^{-} + 2 H X^{+} I X_{2} + 2 H X_{2} O$ ). The iodine liberated needs 22.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The molarity of $H X_{2} O X_{2}$ is:

Thiosulfate = 0.100

\times

22.0 = 2.20 mmol;

I X_{2}

= 1.10 mmol, and 1 mol

H X_{2} O X_{2}

gives 1 mol

I X_{2}

. So

H X_{2} O X_{2}

= 1.10 mmol in 10.0 mL = 0.110 M.