Respuesta :
Answer: The molarity of [tex]Na_2S_2O_3[/tex] is 0.108 M
Explanation:
[tex]KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2[/tex]
[tex]2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI[/tex]
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
[tex]\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol[/tex]
1 mole of [tex]KIO_3[/tex] produces = 3 moles of [tex]I_2[/tex]
[tex]2.8\times 10^{-4}[/tex] moles of [tex]KIO_3[/tex] produces = [tex]\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}[/tex] moles of [tex]I_2[/tex]
Now 1 mole of [tex]I_2[/tex] uses = 2 moles of [tex]Na_2S_2O_3[/tex]
[tex]8.4\times 10^{-4}[/tex] moles of [tex]I_2[/tex] uses = [tex]\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}[/tex] moles of [tex]Na_2S_2O_3[/tex]
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M[/tex]
The molarity of [tex]Na_2S_2O_3[/tex] is 0.108 M
