Use K = 39, I = 127, O = 16. 0.01 mol of KIO3 reacts with excess KI in acid: IO3- + 5I- + 6H+ -> 3I2 + 3H2O. Select the correct statement(s).

The I2 requires 0.06 mol of S2O3^2- for titration (to tetrathionate)

The equivalent mass of KIO3 in this reaction is 42.8

The I2 requires 0.06 mol of S2O3^2- for titration (to tetrathionate)

The equivalent mass of KIO3 in this reaction is 42.8

The mass of I2 liberated is 2.54 g

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3rx159

Use K = 39, I = 127, O = 16. $0.01$ mol of $KIO X_{3}$ reacts with excess $KI$ in acid: $IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O$ . Select the correct statement(s).

A $0.03$ mol of $I X_{2}$ is liberated
BThe $I X_{2}$ requires $0.06$ mol of $S X_{2} O X_{3} X^{2 -}$ for titration (to tetrathionate)
CThe equivalent mass of $KIO X_{3}$ in this reaction is $42.8$
DThe mass of $I X_{2}$ liberated is $2.54 g$

I X_{2} = 3 \times 0.01 = 0.03

mol (A true); thiosulfate

= 2 \times 0.03 = 0.06

mol (B true).

IO X_{3} X^{-}

(

+ 5 \to 0

) has

n = 5

and

M = 214

, so equivalent mass

= 42.8

(C true). Mass of

I X_{2}

= 0.03 \times 254 = 7.62 g

, not

2.54

(D false).