Respuesta :
Answer: 0.22 M
Explanation:
[tex]2S_2O_3^{2-}(aq)+I_3^-\rightarrow S_4O_6^{2-}(aq)+3I^-(aq)[/tex]
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]
moles of [tex]Na_2S_2O_3=Molarity\times {\text {Volume in L}}=0.440\times 0.025=0.011moles[/tex]
[tex]Na_2S_2O_3\rightarrow 2Na^+S_2O_3^{2-}[/tex]
Thus moles of [tex]S_2O_3^{2-}[/tex] = 0.011
According to stoichiometry:
2 moles of [tex]S_2O_3^{2-}(aq)[/tex] require 1 mole of [tex]I_3^[/tex]
Thus 0.011 moles of [tex]S_2O_3^{2-}(aq)[/tex] require=[tex]\frac{1}{2}\times 0.011=5.5\times 10^{-3}[/tex] moles of [tex]I_3^-[/tex]
Thus Molarity of [tex]I_3^-=\frac{5.5\times 10^{-3}}{0.025L}=0.22M[/tex]
Therefore, the molarity of [tex]I_3^-[/tex] in the solution is 0.22 M
