0.1 M solution of KI reacts with excess of ${H}_{2}S{O}_{4}$ and $KI{O}_{3}$ solutions. According to equation

$5{I}^{-}+I{O}_{3}^{-}+6{H}^{+}\to 3{I}_{2}+3{H}_{2}O$

Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of $KI{O}_{3}$
(B) 200 mL of KI solution reacts with 0.006 mol of ${H}_{2}S{O}_{4}$
(C) 0.5 L of KI solution produced 0.005 mol of ${I}_{2}$
(D) Equivalent weight of $KI{O}_{3}$ is equal to ($\frac{\text{ Molecular weight }}{5}$)

Choose the correct answer from the options given below

Question

0.1 M solution of KI reacts with excess of ${H}_{2}S{O}_{4}$ and $KI{O}_{3}$ solutions. According to equation

$5{I}^{-}+I{O}_{3}^{-}+6{H}^{+}\to 3{I}_{2}+3{H}_{2}O$

Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of $KI{O}_{3}$
(B) 200 mL of KI solution reacts with 0.006 mol of ${H}_{2}S{O}_{4}$
(C) 0.5 L of KI solution produced 0.005 mol of ${I}_{2}$
(D) Equivalent weight of $KI{O}_{3}$ is equal to ($\frac{\text{ Molecular weight }}{5}$)

Choose the correct answer from the options given below

YESPYQ · Accepted Answer

Correct answer: b) (A) and (D) only. $200mL(0.2L)$ of KI solution $(0.1M)$ $\to$ Moles of Kl (I – ) $=0.1\times 0.2=0.02\text{ moles }.$ From reaction: 5 moles of ${I}^{-}$react with 1 mole of $I{O}_{3}^{-}$. Required $I{O}_{3}^{-}:0.02\times 1/5=0.004$ moles. Equivalent weight formula: $\text{ Equivalent weight }=\frac{\text{ Molecular weight }}{n-\text{ factor }}$ For $I{O}_{3}^{-}$in this reaction, the change in oxidation state ( n -factor) is 5 . Hence, Equivalent weight =( Molecular weight)/5.

0.1 M solution of KI reacts with excess of \({H}_{2}S{O}_{4}\) and \(KI{O}_{3}\) solutions. According to equation \(5{I}…

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