0.1070 g of pure KIO3 (molar mass 214 g mol^-1) is dissolved, excess KI and acid are added (IO3- + 5I- + 6H+ -> 3I2 + 3H2O) and the iodine liberated needs 29.6 mL of a thiosulphate solution. Find the molarity of the thiosulphate (three decimals).

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JEE Main levelNumerical answerDifficulty 2 of 3eq084

$0.1070 g$ of pure $KIO X_{3}$ (molar mass $214 g mo l^{- 1}$ ) is dissolved, excess $KI$ and acid are added ( $IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O$ ) and the iodine liberated needs $29.6 mL$ of a thiosulphate solution. Find the molarity of the thiosulphate (three decimals).

Numerical answer type, answer in M, tolerance 0.001.

Answer: 0.101 M

mol KIO3 = 0.1070/214 = 5.00e-04; I2 = 3 x 5.00e-04 = 1.50e-03; thiosulphate = 2 x I2 = 3.00e-03 mol. Molarity = 3.00e-03/0.0296 L = 0.1014, about 0.101 M.

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0.1070 g of pure KIOX3​ (molar mass 214 gmol−1) is dissolved, excess KI and acid are added (IOX3​X−+5IX−+6HX+​3IX2​+3HX2​O) and the iodine liberated needs 29.6 mL of a thiosulphate solution. Find the molarity of the thiosulphate (three decimals).

Answer: 0.101 M

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