0.1 M solution of KI reacts with excess of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{KIO}_3$ solutions. According to equation

$$ 5 \mathrm{I}^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O} $$

Identify the correct statements :

(A) 200 mL of KI solution reacts with 0.004 mol of $\mathrm{KIO}_3$

(B) 200 mL of KI solution reacts with 0.006 mol of $\mathrm{H}_2 \mathrm{SO}_4$

(C) 0.5 L of KI solution produced 0.005 mol of $\mathrm{I}_2$

(D) Equivalent weight of $\mathrm{KIO}_3$ is equal to ( $\frac{\text { Molecular weight }}{5}$ )

Choose the correct answer from the options given below :

<p>Molarity of KI = 0.1 M</p> <p>$5{I^ - } + IO_3^ - + 6{H^ + } \to 3{I_2} + 3{H_2}O$ (balanced chemical equation)</p> <p>$5KI + KI{O_3} + 3{H_2}S{O_4} \to 3{I_2} + 3{K_2}S{O_4} + 3{H_2}O$</p> <p>(A) Volume given (KI) = 200 mL = 0.2 L</p> <p>Molarity (KI) = 0.1 M or mol L$^{-1}$</p> <p>The molarity formula is used here to calculate the number of moles of K.I.</p> <p>Molarity, $M = {{number\,of\,moles} \over {volume\,in\,L}}$</p> <p>So, number of moles = Molarity $\times$ Volume in L</p> <p>For the given volume of KI,</p> <p>number of moles = 0.1 mol L$^{-1}$ $\times$ 0.2 L</p> <p>= 0.02 mol</p> <p>The stoichiometric ratio between KI and KIO$_3$ is</p> <p>$KI:KI{O_3}$</p> <p>${I^ - }:IO_3^ -$</p> <p>$5:1$</p> <p>$1:{1 \over 5}$</p> <p>For one mol ${I^ - },{1 \over 5}$ mol $IO_3^ -$ is used.</p> <p>So, for 0.2 mol I$^-$,</p> <p>Moles of $IO_3^ - = 0.02 \times {1 \over 5} = 0.004$ mol</p> <p>The statement (A) is correct.</p> <p>(B) Volume given (KI) = 200 mL = 0.2 L</p> <p>Molarity (KI) = 0.1 M</p> <p>number of moles of KI (I$^-$) = Molarity $\times$ Volume (L)</p> <p>$= 0.1\,mol\,{L^{ - 1}} \times 0.2\,L = 0.02\,mol$</p> <p>The stoichiometric ratio between KE and ${H_2}H{O_4}$ is</p> <p>$KI:{H_2}H{O_4}$</p> <p>$5:3$</p> <p>$1:{3 \over 5}$ {3 moles ${H_2}S{O_4}$ can give 6H$^+$}</p> <p>${I^ - },{3 \over 5}$</p> <p>For one mol ${I^ - },{3 \over 5}$ mol ${H_2}S{O_4}$ is used.</p> <p>So, for 0.02 mol I$^-$,</p> <p>moles of ${H_2}S{O_4} = 0.02 \times {3 \over 5} = 0.012$ mol</p> <p>This statement is not correct.</p> <p>(C) Volume gien (KI) = 0.5 L</p> <p>Molarity (KI) = 0.1 M</p> <p>Number of moles of KI = Molarity $\times$ Volume</p> <p>$= 0.1\,mol\,{L^{ - 1}} \times 0.5\,L = 0.05\,mol$</p> <p>The stoichiometric ratio between I$^-$ and I$_2$ is</p> <p>$KI:{I_2}$</p> <p>${I^ - }:{I_2}$</p> <p>$5:3$</p> <p>$1:{3 \over 5}$</p> <p>For 1 mol ${I^ - },{3 \over 5}$ mol I$_2$ is produced.</p> <p>So, for 0.05 mol, moles of ${I_2} = 0.05 \times {3 \over 5} = 0.03$ mol</p> <p>This statement is not correct.</p> <p>(D) Equivalent weight of $KI{O_3} = {{Molecular\,weight} \over 5}$</p> <p>This statement is correct.</p> <p>Equivalent weight $= {{Molecular\,weight} \over {Valency\,factor\, \to number\,of\,electrons}}$</p> <p>Here, valency factor = 5</p> <p>Each KIO$_3$ molecule gains 5 electrons in this vacation.</p> <p>KIO$_3$ acts as the ordinating agent, gains electrons from the iodide ion in KI.</p> <p>When KIO$_3$ is redued to I$_2$ each molecule gains 5 electrons. So, the valency factor is 5 and equivalent weight is lower than molecular weight.</p> <p>This statement is correct.</p> <p>Correct statements are A and D.</p>