Molarity of KI = 0.1 M

$$5{I^ - } + IO_3^ - + 6{H^ + } \to 3{I_2} + 3{H_2}O$$ (balanced chemical equation)

$$5KI + KI{O_3} + 3{H_2}S{O_4} \to 3{I_2} + 3{K_2}S{O_4} + 3{H_2}O$$

(A) Volume given (KI) = 200 mL = 0.2 L

Molarity (KI) = 0.1 M or mol L$^{-1}$

The molarity formula is used here to calculate the number of moles of K.I.

Molarity, $$M = {{number\,of\,moles} \over {volume\,in\,L}}$$

So, number of moles = Molarity $\times$ Volume in L

For the given volume of KI,

number of moles = 0.1 mol L$^{-1}$ $\times$ 0.2 L

= 0.02 mol

The stoichiometric ratio between KI and KIO$_3$ is

$$KI:KI{O_3}$$

$${I^ - }:IO_3^ - $$

$$5:1$$

$$1:{1 \over 5}$$

For one mol $${I^ - },{1 \over 5}$$ mol $$IO_3^ - $$ is used.

So, for 0.2 mol I$^-$,

Moles of $$IO_3^ - = 0.02 \times {1 \over 5} = 0.004$$ mol

The statement (A) is correct.

(B) Volume given (KI) = 200 mL = 0.2 L

Molarity (KI) = 0.1 M

number of moles of KI (I$^-$) = Molarity $\times$ Volume (L)

$$ = 0.1\,mol\,{L^{ - 1}} \times 0.2\,L = 0.02\,mol$$

The stoichiometric ratio between KE and $${H_2}H{O_4}$$ is

$$KI:{H_2}H{O_4}$$

$$5:3$$

$$1:{3 \over 5}$$ {3 moles $${H_2}S{O_4}$$ can give 6H$^+$}

$${I^ - },{3 \over 5}$$

For one mol $${I^ - },{3 \over 5}$$ mol $${H_2}S{O_4}$$ is used.

So, for 0.02 mol I$^-$,

moles of $${H_2}S{O_4} = 0.02 \times {3 \over 5} = 0.012$$ mol

This statement is not correct.

(C) Volume gien (KI) = 0.5 L

Molarity (KI) = 0.1 M

Number of moles of KI = Molarity $\times$ Volume

$$ = 0.1\,mol\,{L^{ - 1}} \times 0.5\,L = 0.05\,mol$$

The stoichiometric ratio between I$^-$ and I$_2$ is

$$KI:{I_2}$$

$${I^ - }:{I_2}$$

$$5:3$$

$$1:{3 \over 5}$$

For 1 mol $${I^ - },{3 \over 5}$$ mol I$_2$ is produced.

So, for 0.05 mol, moles of $${I_2} = 0.05 \times {3 \over 5} = 0.03$$ mol

This statement is not correct.

(D) Equivalent weight of $$KI{O_3} = {{Molecular\,weight} \over 5}$$

This statement is correct.

Equivalent weight $$ = {{Molecular\,weight} \over {Valency\,factor\, \to number\,of\,electrons}}$$

Here, valency factor = 5

Each KIO$_3$ molecule gains 5 electrons in this vacation.

KIO$_3$ acts as the ordinating agent, gains electrons from the iodide ion in KI.

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.

This statement is correct.

Correct statements are A and D.