Select the correct statements (iodometry with excess KI; masses: Cu 63.5, S 32, O 16, H 1).

AThe equivalent mass of $CuSO X_{4} \cdot 5 H X_{2} O$ (249.5 g/mol) is 124.75 g/eq
BIn $2 S X_{2} O X_{3} X^{2 -} S X_{4} O X_{6} X^{2 -} + 2 e X^{-}$ , the n-factor of thiosulfate is 1
C0.1 mol of $KIO X_{3}$ liberates 0.3 mol of $I X_{2}$ from excess acidified KI
D1 mol of $K X_{2} Cr X_{2} O X_{7}$ liberates 3 mol of $I X_{2}$ from excess acidified KI

Answer: B, C, D

CuSO X_{4} \cdot 5 H X_{2} O

gains 1 e

^{-}

per

Cu X^{2 +}

(n = 1), so equivalent mass is 249.5, not 124.75 (false). Two thiosulfate lose 2 e

^{-}

in total, so n = 1 (true).

IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O

gives 3 mol

I X_{2}

per mol (true), so 0.3 mol from 0.1 mol.

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

gains 6 e

^{-}

, giving 3

I X_{2}

(true).