Answer: 0.67 g
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]
[tex]moles of [tex]Na_2S_2O_3=Molarity\times {\text {Volume in L}}=0.150\times 0.0352=5.28\times 10^{-3}moles[/tex]
[tex]I_2(aq)+2S_2O_3^{2-}(aq) S_4O_6^{2-}(aq)+2I^-(aq)[/tex]
According to stoichiometry:
2 moles of [tex]2S_2O_3^{2-}(aq)[/tex] require 1 mole of [tex]I_2[/tex]
Thus [tex]5.28\times 10^{-3}moles[/tex] require=[tex]\frac{1}{2}\times 5.28\times10^{-3}=2.64\times 10^{-3}[/tex] moles of [tex]I_2[/tex]
Mass of [tex]I_2=moles\times {\text {molar mass}}=2.64\times 10^{-3}\times 254=0.67g[/tex]
Thus 0.67 g of iodine are present in the solution.
