JEE Main 2025 Jan 29 Shift 2, Chemistry Q9: Advance Stoichiometry Based on Redox Reactions

JEE Main2025Jan 29, Shift 2Chemistry

0.1 M solution of KI reacts with excess of $H_{2} S O_{4}$ and $K I O_{3}$ solution. According to equation

$5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O$

Identify the correct statements :

(A) 200 mL of KI solution reacts with 0.004 mol of $K I O_{3}$

(B) 200 mL of KI solution reacts with 0.006 mol of $H_{2} S O_{4}$

(D) Equivalent weight of $K I O_{3}$ is equal to $(\frac{Molecular weight}{5})$

Choose the correct answer from the options given below:

A
(A) and (D) only
B
(B) and (C) only
C
(A) and (B) only
D
(C) and (D) only

Solution

Moles of $K I$ in 200 mL of 0.1 M solution $= 0.2 \times 0.1 = 0.02$ mol $= n (I^{-})$ .

(A) Stoichiometry: $5 I^{-} : 1 I O_{3}^{-}$ , so $n (K I O_{3}) = 0.02/5 = 0.004$ mol. Correct.

(B) $H^{+}$ required $= 0.02 \times 6/5 = 0.024$ mol, so $H_{2} S O_{4} = 0.024/2 = 0.012$ mol (not $0.006$ ). Incorrect.

(C) In 0.5 L of 0.1 M $K I$ , $n (I^{-}) = 0.05$ mol. $I_{2}$ produced $= 0.05 \times 3/5 = 0.03$ mol (not $0.005$ ). Incorrect.

(D) In $I O_{3}^{-}$ iodine is at $+ 5$ oxidation state and is reduced to $0$ in $I_{2}$ , a $5$ -electron change per iodine. Hence $n$ -factor of $K I O_{3} = 5$ and equivalent weight $= M /5$ . Correct.

$∴$ (A) and (D) are correct.

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JEE Main 2025 Jan 29 Shift 2, Chemistry Q9: Advance Stoichiometry Based on Redox Reactions

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