Use K = 39, Mn = 55, O = 16. 2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O is carried out with 3.16 g of KMnO4 and excess HCl, with 85% of the theoretical Cl2 collected. The Cl2 is passed into excess KI solution (Cl2 + 2I- -> 2Cl- + I2) and the I2 is titrated with 0.50 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-). Find the volume of thiosulfate in mL (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3rx155

Use K = 39, Mn = 55, O = 16. $2 KMnO X_{4} + 16 HCl 2 KCl + 2 MnCl X_{2} + 5 Cl X_{2} + 8 H X_{2} O$ is carried out with $3.16 g$ of $KMnO X_{4}$ and excess $HCl$ , with $85%$ of the theoretical $Cl X_{2}$ collected. The $Cl X_{2}$ is passed into excess $KI$ solution ( $Cl X_{2} + 2 I X^{-} 2 Cl X^{-} + I X_{2}$ ) and the $I X_{2}$ is titrated with $0.50 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). Find the volume of thiosulfate in mL (nearest integer).

Numerical answer type, answer in mL, tolerance 0.5.

n (KMnO X_{4}) = 3.16/158 = 0.02

Cl X_{2}

(theoretical)

= 0.02 \times 5/2 = 0.05

mol; collected

= 0.05 \times 0.85 = 0.0425

mol. Thiosulfate

= 2 \times 0.0425 = 0.085

mol.

V = 0.085/0.50 = 170 mL