0.2646 g of pure $K X_{2} Cr X_{2} O X_{7}$ (294 g/mol) is dissolved, acidified, and treated with excess KI. The iodine liberated requires 28.0 mL of sodium thiosulfate solution. The molarity of the thiosulfate is:

Answer: D

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

, so 1 mol dichromate gives 3

I X_{2}

, which need 6 thiosulfate. Dichromate = 0.2646/294 = 0.900 mmol, thiosulfate = 5.40 mmol. Molarity = 5.40/28.0 = 0.193 M.