Use Cu = 63.5, S = 32. In acidic solution KMnO4 (MnO4- -> Mn^2+) oxidises Cu2S to Cu^2+ and SO2. The volume of 0.04 M KMnO4 needed for 1.59 g of Cu2S is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx151

Use Cu = 63.5, S = 32. In acidic solution $KMnO X_{4}$ ( $MnO X_{4} X^{-} Mn X^{2 +}$ ) oxidises $Cu X_{2} S$ to $Cu X^{2 +}$ and $SO X_{2}$ . The volume of $0.04 M$ $KMnO X_{4}$ needed for $1.59 g$ of $Cu X_{2} S$ is:

M (Cu X_{2} S) = 159

, so moles

= 0.01

n

-factor: two

Cu X^{+} Cu X^{2 +}

give 2 e

^{-}

;

S X^{2 -} S X^{4 +}

gives 6 e

^{-}

; total

8

. Equivalents

= 0.08

KMnO X_{4}

moles

= 0.08/5 = 0.016

V = 0.016/0.04 = 0.4 L = 400 mL