0.490 g of K2Cr2O7 (molar mass 294 g mol^-1) is treated with excess KI in dilute H2SO4 and the liberated iodine is titrated with 0.100 M Na2S2O3 (I2 + 2S2O3^2- -> 2I^- + S4O6^2-). The volume of thiosulfate required is

JEE Advanced levelSingle correct optionDifficulty 3 of 3df103

$0.490 g$ of $K X_{2} Cr X_{2} O X_{7}$ (molar mass $294 g mo l^{- 1}$ ) is treated with excess KI in dilute $H X_{2} SO X_{4}$ and the liberated iodine is titrated with $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} \to 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). The volume of thiosulfate required is

Cr X_{2} O X_{7} X^{2 -} + 6 I X^{-} + 14 H X^{+} \to 2 Cr X^{3 +} + 3 I X_{2} + 7 H X_{2} O

, and each

I X_{2}

needs 2

S X_{2} O X_{3} X^{2 -}

, so 1 mol dichromate corresponds to 6 mol thiosulfate. Moles of dichromate

= 0.490/294 = 1.667 \times 1 0^{- 3}

, so thiosulfate

= 0.0100 mol

, and

V = 0.0100/0.100 = 0.100 L = 100.0 mL