In the process ${\mathrm{I}}_2+{\mathrm{I}}^{-}\rightleftharpoons {{\mathrm{I}}_3}^{-}$(in aq medium), initially there are 2 mole ${\mathrm{I}}_2$ & 2 mole ${\mathrm{I}}^{-}$. But at equilibrium, due to addition of ${\mathrm{AgNO}}_{3\left(\mathrm{aq}\right)},1.75$ mole yellow ppt is obtained. ${\mathrm{K}}_{\mathrm{C}}$ for the process is $\left({\mathrm{V}}_{\text{flask }}=1{\mathrm{dm}}^3\right)$ nearly

 The given reaction,

${\mathrm{I}}_2+{\mathrm{I}}^{-}\rightleftarrows {\mathrm{I}}_3^{-}$

2     2

(2-x) (2-x)   x

Since, Moles of $\mathrm{AgI}=$ Moles of ${\mathrm{I}}^{-}$in solution

$\therefore {\left[{\mathrm{I}}^{-}\right]}_{\mathrm{eqm}}=1.75\mathrm{M}$ 

$2-\mathrm{x}=2$

$\therefore \mathrm{x}=1$

Equilibrium constant of the reaction at equilibrium,

$\mathrm{K}$ =$\frac{\mathrm{X}}{(2-\mathrm{x})(2-\mathrm{x})}$

$\mathrm{K}$=$\frac{1}{0.25\times 0.25}=16$

Given that,

${\mathrm{V}}_{\text{flask }}=1{\mathrm{dm}}^3$

$\mathrm{K}=\frac{16}{100}=0.16$

Hence the correct answer is (C) $0.16$