A sodium thiosulphate solution requires 20.0 mL of 0.0500 M $I X_{2}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). An equal volume of the same thiosulphate solution is oxidised completely to sulphate by chlorine water ( $Cl X_{2} 2 Cl X^{-}$ ). The volume of 0.0500 M $Cl X_{2}$ solution required is:

Answer: A

With iodine:

I X_{2}

= 1.00 mmol, so thiosulphate = 2.00 mmol. With chlorine, each

S X_{2} O X_{3} X^{2 -}

(S: +2 to +6) loses 8 electrons, so 16 meq. Each

Cl X_{2}

takes 2 electrons, so mmol

Cl X_{2}

= 16/2 = 8.00. Volume = 8.00/0.0500 = 160 mL.