0.1 M solution of KI reacts with excess of H 2 SO 4 and KIO 3 solutions. According to equation 5 I − + IO 3 − + 6 H + → 3 I 2 + 3 H 2 O Identify the correct statements : (A) 200 mL of KI s

Question

0.1 M solution of KI reacts with excess of $H_{2} SO_{4}$ and $KIO_{3}$ solutions. According to equation
$5 I^{-} + IO_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O$

Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of $KIO_{3}$
(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 :

Options:

Answer:

A

Solution:

$E_{KIO_{3}} = \frac{Molecular welght}{n _{f}} n_{f} = 5 E_{KIO_{3}} = \frac{Molecular welght}{5}$
(D) is correct meq of $KI = 0.1 \times 200 = 20$ meq of $KIO_{3} = 4 \times 5 = 20$
$(A)$ is correct

TestHub · Accepted Answer

​ E KIO 3 ​ ​ = n f ​ Molecular welght ​ n f ​ = 5 E KIO 3 ​ ​ = 5 Molecular welght ​ ​ (D) is correct meq of KI = 0.1 × 200 = 20 meq of KIO 3 ​ = 4 × 5 = 20 ( A ) is correct

Chemistry - Redox Reactions Question with Solution | TestHub

Options:

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