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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2025 · 29 January · Evening Shift · Chemistry · Redox Reactions

0.1 M solution of KI reacts with excess of H_2 SO_4 and KIO_3...

0.1 M solution of KI reacts with excess of H2SO4\mathrm{H}_2 \mathrm{SO}_4 and KIO3\mathrm{KIO}_3 solutions. According to equation 55\mathrm{I}−+IO3−+6H+→3^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3\mathrm{I}2+3H2O_2+3 \mathrm{H}_2 \mathrm{O} Identify the correct statements : (A) 200 mL of KI solution reacts with 0.004 mol of KIO3\mathrm{KIO}_3 (B) 200 mL of KI solution reacts with 0.006 mol of H2SO4\mathrm{H}_2 \mathrm{SO}_4 (C) 0.5 L of KI solution produced 0.005 mol of I2\mathrm{I}_2 (D) Equivalent weight of KIO3\mathrm{KIO}_3 is equal to (  Molecular weight 5\frac{\text{ Molecular weight }}{5} ) Choose the correct answer from the options given below :

Options

  1. A(A) and (D) only
  2. B(C) and (D) only
  3. C(B) and (C) only
  4. D(A) and (B) only

Correct answer

A. (A) and (D) only

Step-by-step solution

EKIO3= Molecular welght nfnf=5EKIO3= Molecular welght 5\begin{aligned} & \mathrm{E}_{\mathrm{KIO}_3}=\frac{\text{ Molecular welght }}{\mathrm{n}_{\mathrm{f}}} \\ & \mathrm{n}_{\mathrm{f}}=5 \\ & \mathrm{E}_{\mathrm{KIO}_3}=\frac{\text{ Molecular welght }}{5}\end{aligned} (D) is correct meq of KI=0.1×200=20\mathrm{KI}=0.1 \times 200=20 meq of KIO3=4×5=20(A)\mathrm{KIO}_3=4 \times 5=20 (\mathrm{A}) is correct
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