0.1 M solution of reacts with excess of and solutions. According to equation Identify the correct statements: (A) 200 mL of KI solution reacts with 0.004 mol of (B) 200 mL of KI solution reacts with 0.006 mol of (C) 0.5 L of KI solution produced

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Solution

Q.
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: [2025]

1 (A) and (B) only

2 (A) and (D) only

3 (B) and (C) only

4 (C) and (D) only

Ans.
(2)

(A) $Moles of KI = Molarity \times Volume in L = 0.1 M \times 0.2 L = 0.02 mol$

As per reaction stoichiometry:

$\frac{n_{{KIO}_{3}}}{1} = \frac{n_{KI}}{5}$

$\frac{n_{{KIO}_{3}}}{1} = \frac{0.02}{5}$

$n_{{KIO}_{3}} = 0.004 mol$

(B) $Moles of KI = Molarity \times Volume in L$

$=0.1 M×0.2 L = 0.02 mol$

As per reaction stoichiometry:

$\frac{n_{H^{+}}}{6} = \frac{n_{KI}}{5}$

$\frac{2 \times n_{H_{2} {SO}_{4}}}{6} = \frac{n_{KI}}{5}$

$\frac{2 \times n_{H_{2} {SO}_{4}}}{6} = \frac{0.02}{5}$

$n_{H_{2} {SO}_{4}} = 0.012 mol$

(C) $Moles of KI = Molarity \times Volume in L$

$= 0.1 M×0.5 L = 0.05 mol$

As per reaction stoichiometry:

$\frac{n_{I_{2}}}{3} = \frac{n_{KI}}{5}$

$\frac{n_{I_{2}}}{3} = \frac{0.5}{5}$

$n_{I_{2}} = 0.50 mol$

(D) $n {factor of KIO}_{3} = ({Change in O.N. of I from KIO}_{3} {to I}_{2}) \times {Number of I in KIO}_{3} = (5) \times 1 = 5$

${Equivalent weight of KIO}_{3} = \frac{{Molecular weight of KIO}_{3}}{n factor} = \frac{{Molecular weight of KIO}_{3}}{5}$

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