0.1 M solution of KI reacts with excess of H₂ SO₄ and KIO₃ solutions. According to equation
5 I⁻+IO₃⁻+6 H⁺ → 3 I₂+3 H₂ O
Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of KIO₃
(B) 200 mL of KI solution reacts with 0.006 mol of H₂ SO₄
(C) 0.5 L of KI solution produced 0.005 mol of I₂
(D) Equivalent weight of KIO₃ is equal to ( (Molecular weight)/5 )
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Convert the KI into moles of I⁻ first, then let the 5 : 1 : 6 : 3 coefficients test each statement.

1. Moles of iodide. Moles of I⁻ = molarity × volume in litres. So 200 mL gives 0.1 × 0.2 = 0.02 mol, and 0.5 L gives 0.1 × 0.5 = 0.05 mol.
2. Read the ratios. For every 5 mol I⁻, the equation uses 1 mol IO₃⁻ and 6 mol H⁺, and gives 3 mol I₂. Each H₂SO₄ supplies 2 H⁺. In IO₃⁻, iodine goes from +5 to 0, so it gains 5 electrons and its n-factor is 5.

   Statement	Working	Correct value	Verdict
   (A) 0.004 mol KIO₃	0.02/5	0.004 mol	correct
   (B) 0.006 mol H₂SO₄	H⁺ = 6 × 0.004 = 0.024, so H₂SO₄ = 0.024/2	0.012 mol	wrong
   (C) 0.005 mol I₂	3/5 × 0.05	0.03 mol	wrong
   (D) Eq. wt. = M/5	I: +5 to 0, n-factor 5	M/5	correct

3. So. Only (A) and (D) are correct, which is option (d).

The idea to keep. In stoichiometry, get moles of the known species first, then scale by the coefficients. Change H⁺ into acid moles using its basicity, and find equivalent weight from the electrons each formula unit gains or loses.

Answer from the NTA final answer key. Solution by PrepWiser.