25.0 mL of a $CuSO X_{4}$ solution is treated with excess KI and the liberated iodine needs 20.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The molarity of $Cu X^{2 +}$ in the solution is:

Answer: C

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

and

I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}

give 1 mol

Cu X^{2 +}

\equiv

1 mol thiosulfate. Thiosulfate = 0.100

\times

20.0 = 2.00 mmol, so

Cu X^{2 +}

= 2.00 mmol in 25.0 mL = 0.0800 M.