0.1 M solution of KI reacts with excess of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{KIO}_3$ solutions.…

0.1 M solution of KI reacts with excess of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{KIO}_3$ solutions. According to equation
$5 \mathrm{I}^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}$
Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of $\mathrm{KIO}_3$
(B) 200 mL of KI solution reacts with 0.006 mol of $\mathrm{H}_2 \mathrm{SO}_4$
(C) 0.5 L of KI solution produced 0.005 mol of $\mathrm{I}_2$
(D) Equivalent weight of $\mathrm{KIO}_3$ is equal to ( $\frac{\text { Molecular weight }}{5}$)
Choose the correct answer from the options given below :
  1. (A) and (D) only
  2. (C) and (D) only
  3. (B) and (C) only
  4. (A) and (B) only

Solution

$\begin{aligned} & \mathrm{E}_{\mathrm{KIO}_3}=\frac{\text { Molecular welght }}{\mathrm{n}_{\mathrm{f}}} \\ & \mathrm{nff}_{\mathrm{f}}=5 \\ & \mathrm{E}_{\mathrm{KIO}_3}=\frac{\text { Molecular welght }}{5}\end{aligned}$
(D) is correct meq of $\mathrm{KI}=0.1 \times 200=20$ meq of $\mathrm{KIO}_3=4 \times 5=20$
$(\mathrm{A})$ is correct

Asked in: JEE Main 2025 (29 Jan Shift 2)

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