$25.0 mL$ of $0.0500 M$ $K X_{2} Cr X_{2} O X_{7}$ is acidified and treated with excess $KI$ . The iodine liberated is titrated with $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ (oxidised to $S X_{4} O X_{6} X^{2 -}$ ). The volume of thiosulphate required is:

Answer: D

Cr X_{2} O X_{7} X^{2 -} + 6 I X^{-} + 14 H X^{+} 2 Cr X^{3 +} + 3 I X_{2} + 7 H X_{2} O

, and each

I X_{2}

reacts with 2

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

, so 1 mol dichromate corresponds to 6 mol thiosulphate. Mmol of dichromate

= 25.0 \times 0.0500 = 1.25

; thiosulphate

= 7.50

mmol. Volume

= 7.50/0.100 = 75.0 mL