A thiosulphate solution is standardised with KIO3 (molar mass 214). 0.2140 g of KIO3 reacts with excess KI in acid (IO3- + 5I- + 6H+ -> 3I2 + 3H2O) and the iodine needs 36.0 mL of the thiosulphate solution. The molarity of the thiosulphate solution is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx207

A thiosulphate solution is standardised with $KIO X_{3}$ (molar mass 214). 0.2140 g of $KIO X_{3}$ reacts with excess KI in acid ( $IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O$ ) and the iodine needs 36.0 mL of the thiosulphate solution. The molarity of the thiosulphate solution is:

mmol

KIO X_{3}

= 0.2140/214 x 1000 = 1.000. Iodine formed = 3.000 mmol, which needs 6.000 mmol thiosulphate. Molarity = 6.000/36.0 = 0.1667 M.