0.500 g of a brass sample is dissolved and the copper determined iodometrically with excess KI; the iodine liberated needs 24.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ (Cu = 63.5). The percentage of Cu in the brass is:

Answer: C

2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}

;

I X_{2}

then takes 2 thiosulfate, so 1 mol Cu

\equiv

1 mol thiosulfate. Cu = 0.100

\times

24.0 = 2.40 mmol = 152.4 mg. Percentage = 0.1524/0.500

\times

100 = 30.5 %.