20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide…

$\displaystyle 20$ mL of sodium iodide solution gave $\displaystyle 4.74$ g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is $\displaystyle \_\_\_\_$ M. (Nearest Integer value) (Given : $\displaystyle \mathrm{Na}=23, \mathrm{I}=127, \mathrm{Ag}=108, \mathrm{~N}=14, \mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ )