$0.1\mathrm{M}$ solution of $\mathrm{KI}$ reacts with excess of $H_{2}S{O}_4$and $KI{O}_3$ solutions, according to equation $5I^{-}+I{O}_3^{-}+6H^{+}⟶3I_2+3H_{2}O$; which of the following statements is/are correct :

$0.1\mathrm{M}$ solution of $\mathrm{KI}$ reacts with excess of $H_{2}S{O}_4$and $KI{O}_3$ solutions, according to equation : $5I^{-}+I{O}_3^{-}+6H^{+}⟶3I_2+3H_{2}O$

(a) 200 mL of the $\mathrm{KI}$ solution react with 0.004 mole $KI{O}_3$

$=200\times 0.1$

$=\frac{ 200 \times 0.1}{5}$

(b) 100 mL of the $\mathrm{KI}$ solution reacts with 0.006 mole of$H_{2}S{O}_4$.

$=100\times 0.1$

$$\begin{gathered} 100\times 0.1\frac{6}{5} \\ \Rightarrow n_{{\mathrm{H}}_2{\mathrm{SO}}_4} \end{gathered}$$

$\begin{array}{l}=\frac{100\times 0.1\times 6}{1000\times 2\times 5}\\ =0.006\end{array}$

(d) Equivalent weight of $KI{O}_3$ is equal to $\frac{( \mathrm{Molecular} \mathrm{weight} )}{5}$.
 

$E_{{\mathrm{KIO}}_3}=\frac{M}{5}$

Hence, the correct option is a, b and d.