A 25.0 mL portion of a K2Cr2O7 solution is acidified and treated with excess KI (Cr2O7^2- + 6I- + 14H+ -> 2Cr^3+ + 3I2 + 7H2O). The liberated I2 needs 27.4 mL of 0.100 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-). Find the concentration of the K2Cr2O7 solution in mmol/L (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3rx154

A $25.0 mL$ portion of a $K X_{2} Cr X_{2} O X_{7}$ solution is acidified and treated with excess $KI$ ( $Cr X_{2} O X_{7} X^{2 -} + 6 I X^{-} + 14 H X^{+} 2 Cr X^{3 +} + 3 I X_{2} + 7 H X_{2} O$ ). The liberated $I X_{2}$ needs $27.4 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). Find the concentration of the $K X_{2} Cr X_{2} O X_{7}$ solution in $mmol/L$ (two decimals).

Numerical answer type, answer in mmol/L, tolerance 0.03.

Thiosulfate

= 27.4 \times 0.100 = 2.74

mmol. Each

Cr X_{2} O X_{7} X^{2 -}

gives 3

I X_{2}

, i.e. needs 6 thiosulfate, so dichromate

= 2.74/6 = 0.4567

mmol in

25.0 mL

. Concentration

= 0.4567/25.0 \times 1000 = 18.27 mmol/L