0.1 M solution of KI reacts with excess of $$H_{2}SO_{4}$$ and $$KIO{3}$$ solutions. According to equation $$5I^- + IO_3^- + 6H^+ \rightarrow 3I_2 + 3H_2O$$ Identify the correct statements : (A) 200 mL of KI solution reacts with 0.004 mol of $$KIO_{3}$$ (B) 200 mL of KI solution reacts with 0.006 mol of $$H_{2}SO_{4}$$ (C) 0.5 L of KI solution produced 0.005 mol of $$I_{2}$$ (D) Equivalent weight of $$KIO_{3}$$ is equal to ( $$\frac{Molecular weight}{5}$$ ) Choose the correct answer from the options given below :

The reaction: $$5I^- + IO_3^- + 6H^+ \to 3I_2 + 3H_2O$$

0.1 M KI solution: moles of $$I^-$$ in 200 mL = $$0.1 \times 0.2 = 0.02$$ mol.

(A) 200 mL of KI has 0.02 mol $$I^-$$. From stoichiometry, 5 mol $$I^-$$ reacts with 1 mol $$IO_3^-$$.

Moles of $$KIO_3 = 0.02/5 = 0.004$$ mol. ✓

(B) 200 mL has 0.02 mol $$I^-$$. From stoichiometry, 5 mol $$I^-$$ reacts with 6 mol $$H^+$$.

Moles of $$H^+ = 0.02 \times 6/5 = 0.024$$ mol. Moles of $$H_2SO_4 = 0.024/2 = 0.012$$ mol ≠ 0.006. ✗

(C) 0.5 L of KI has $$0.1 \times 0.5 = 0.05$$ mol $$I^-$$. From stoichiometry, 5 mol $$I^-$$ produces 3 mol $$I_2$$.

Moles of $$I_2 = 0.05 \times 3/5 = 0.03$$ mol ≠ 0.005. ✗

(D) In this reaction, $$IO_3^-$$ goes from I(+5) to I₂(0), gaining 5 electrons per I atom.

Equivalent weight = Molecular weight / 5. ✓

The correct answer is Option 1: (A) and (D) only.