1.00 g of a chromium steel is treated so that all its chromium is converted to dichromate. After adding excess KI in acid, the liberated iodine needs 21.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . Find the percentage of Cr in the steel (Cr = 52), to one decimal place.

Numerical answer type, answer in %, tolerance 0.05.

Answer: 3.6 %

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

and

I X_{2} \equiv 2 S X_{2} O X_{3} X^{2 -}

, so 1 dichromate (2 Cr) gives 6 thiosulphate: 1 Cr is equivalent to 3 thiosulphate. mmol thiosulphate = 2.10, so mmol Cr = 0.700. Mass =

0.700 \times 52 = 36.4

mg, i.e. 3.64% of 1.00 g.