Question

0.1 M solution of KI reacts with excess of \( \text{H}_2\text{SO}_4 \) and KIO\(_3\), according to the equation: \[ 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O \] 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: \[ \frac{\text{Molecular weight}}{5} \] Choose the correct answer from the options given below:

Hint:

The equivalent weight of a substance is defined as its molecular weight divided by the number of electrons involved in the reaction. For KIO\(_3\), 5 moles of electrons are involved per mole of KIO\(_3\).

Answer 1

The Correct Option is A

- Statement (A) is correct because 200 mL of 0.1 M KI contains 0.02 moles of KI, and according to the equation, 5 moles of I\(^-\) react with 1 mole of KIO\(_3\). Therefore, 0.02 moles of KI would require 0.004 mol of KIO\(_3\).
- Statement (C) is also correct because 0.5 L of 0.1 M KI will contain 0.05 moles of KI, and according to the equation, this will produce 0.005 mol of I\(_2\).
- Statement (D) is correct because the equivalent weight of KIO\(_3\) is equal to its molecular weight divided by 5, as 5 moles of iodide react with one mole of KIO\(_3\).
Thus, the correct answer is \( \boxed{(A),(D)} \).