JEE Main 2025 — 29 January, Evening Shift — Question 6

Balanced reaction, $5\,\mathrm{I^-} + \mathrm{IO_3^-} + 6\,\mathrm{H^+} \xrightarrow{} 3\,\mathrm{I_2} + 3\,\mathrm{H_2O}$

Moles of KI in $200\,\mathrm{mL}$ $= 0.1 \times 0.200 = 0.020\,\mathrm{mol}$

Stoichiometry, $5\,\mathrm{I^-}$ reacts with $1\,\mathrm{IO_3^-}$

Moles of $\mathrm{IO_3^-}$ required $= \dfrac{0.020}{5} = 0.004\,\mathrm{mol}$

Moles of $\mathrm{H^+}$ required $= 6\,\mathrm{H^+}$ per $\mathrm{IO_3^-}$

In $0.004\,\mathrm{mol}$ $\mathrm{IO_3^-}$ $= 6 \times 0.004 = 0.024\,\mathrm{mol}$ $\mathrm{H^+}$

$\mathrm{H_2SO_4}$ gives $2\,\mathrm{H^+}$, so moles of $\mathrm{H_2SO_4}$ $= \dfrac{0.024}{2} = 0.012\,\mathrm{mol}$

Therefore, Statement (A) = true Statement (B) = false

For $0.5\,\mathrm{L}$ KI $= 0.1 \times 0.5 = 0.05\,\mathrm{mol}$ $\mathrm{I^-}$

Moles of $\mathrm{I_2}$ produced $= \dfrac{3}{5} \times 0.05 = 0.03\,\mathrm{mol}$

$\Rightarrow$ statement (C) = false

Equivalent weight of $\mathrm{KIO_3}$: $n$ factor = $5$ $\Rightarrow$ $\dfrac{\mathrm{Molecular\ weight}}{5}$

$\Rightarrow$ statement (D) = true

Thus, the correct option is A.