0.1 M solution of KI reacts with excess of H2SO4 and KIO3 solution. According to equation 5I- + IO3- + 6H+ - 3I2 + 3H2O…

Given: 0.1 M KI solution; balanced equation: $\ce{5I^- + IO3^- + 6H^+ -> 3I2 + 3H2O}$

Step 1 — Moles of KI in 200 mL

$n(\ce{KI}) = 0.1 \times 0.200 = 0.02$ mol = 0.02 mol $\ce{I^-}$

Check Statement A: Moles of $\ce{KIO3}$ required

From stoichiometry: $5\text{ mol }\ce{I^-}$ reacts with $1\text{ mol }\ce{IO3^-}$

$n(\ce{IO3^-}) = \frac{0.02}{5} = 0.004$ mol ✓

→ Statement A is CORRECT ✓

Check Statement B: Moles of $\ce{H2SO4}$ required

$6\text{ mol }\ce{H^+}$ per mol $\ce{IO3^-}$; $\ce{H2SO4}$ gives 2 $\ce{H^+}$

$n(\ce{H^+}) = 6 \times 0.004 = 0.024$ mol

$n(\ce{H2SO4}) = \frac{0.024}{2} = 0.012$ mol ≠ 0.008 mol ✗

→ Statement B is INCORRECT ✗

Check Statement C: $\ce{I2}$ from 0.5 L of 0.1 M KI

$n(\ce{I^-}) = 0.1 \times 0.5 = 0.05$ mol

$n(\ce{I2}) = \frac{3}{5} \times 0.05 = 0.03$ mol ≠ 0.005 mol ✗

→ Statement C is INCORRECT ✗

Check Statement D: Equivalent weight of $\ce{KIO3}$

In this reaction, $\ce{I}$ in $\ce{IO3^-}$ goes from +5 to 0 (in $\ce{I2}$): change = 5 electrons per I atom.

Equivalent weight $= \frac{M(\ce{KIO3})}{5} = \frac{214}{5}$ ✓

→ Statement D is CORRECT ✓

Answer: (A) and (D) only → Option (a)