100" " mL of $H_2{O}_2$ solution having volume strength $11.35 \mathrm{V}$is mixed with 50 mL of $0.5\mathrm{M} \mathrm{KI}$ solution to liberate $I_2$ gas. All the $I_2$ gas liberated is trapped to form a 500 mL solution termed as X.200 mL of the solution X of $I_2$ required 50 mL hypo solution for conversion to  $I^{-}$and $S_4{O}_6^{(}2-)$. Assuming all reactions to undergo 100% completion, identify the correct option(s)

Molarity of  ${\mathrm{H}}_2{\mathrm{O}}_2=1$

${\mathrm{H}}_2{\mathrm{O}}_2⟶{\mathrm{H}}_2\mathrm{O}$ and $\mathrm{KI}⟶{\mathrm{I}}_2$

No. of milli equivalent of$H_2{O}_2$ taken $$\begin{gathered} =100\times 1\times 2 \\ =200 \end{gathered}$$

No. of milli equivalent of $\mathrm{KI}$ taken $$\begin{gathered} =50\times 0.5\times 1 \\ =25 \end{gathered}$$

$=$ No. of milli equivalent of $I_2$

No. of milli equivalent of$H_2{O}_2$left $=175$

(a) Vol. strength of remaining $H_2{O}_2$

$\begin{array}{l}=11.35\times \frac{175}{150}\times \frac{1}{2}\\ =6.62\mathrm{V}\end{array}$

(b) Calculate the molarity  

$$\begin{gathered} 500\times M\times 2=25 \\ \Rightarrow M=0.025 \end{gathered}$$

(c) 200 ml of x contains$\to \frac{25}{500}\times 200$ milli equivalent of $I_2$

= milli equivalent of hypo

$\frac{25}{500}\times 200=50\times M\times 1$

$\Rightarrow M=0.2$

(d) Moles of tetrathionate ions formed will be 0.0

$\begin{array}{l}{\mathrm{I}}_2+2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3⟶{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}\\ 50\times 0.2\\ =10\mathrm{m} \mathrm{mol}5\mathrm{m} \mathrm{mol}\end{array}$