0.690 g of pure K2Cr2O7 (molar mass 294 g mol^-1) is treated with excess acidified KI, liberating I2 (dichromate goes to Cr^3+). The iodine is titrated with 0.0800 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-). Volume of thiosulfate needed, in mL (nearest integer)?

JEE Advanced levelNumerical answerDifficulty 3 of 3eq033

$0.690 g$ of pure $K X_{2} Cr X_{2} O X_{7}$ (molar mass $294 g mo l^{- 1}$ ) is treated with excess acidified $KI$ , liberating $I X_{2}$ (dichromate goes to $Cr X^{3 +}$ ). The iodine is titrated with $0.0800 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). Volume of thiosulfate needed, in $mL$ (nearest integer)?

Numerical answer type, answer in mL, tolerance 0.5.

Dichromate has

n = 6

, so equivalents of liberated

I X_{2}

(and hence of thiosulfate)

= 6 \times 0.690/294 = 0.01408

. Thiosulfate here has

n = 1

, so

N = 0.0800 N

. Volume

= 0.01408/0.0800 = 0.1760 L = 176 mL