Answer :
Answer: The molarity of LiOH is 0.139 M
Explanation:
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] .....(1)
Molarity of iron (II) chloride solution = 0.210 M
Volume of solution = 11.6 mL = 0.0116 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.210M=\frac{\text{Moles of }FeCl_2}{0.0116L}\\\\\text{Moles of }FeCl_2=(0.210mol/L\times 0.0116L)=2.44\times 10^{-3}mol[/tex]
For the given chemical reaction:
[tex]FeCl_2(aq.)+2LiOH(aq.)\rightarrow Fe(OH)_2(s)+2LiCl(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of iron (II) chloride reacts with 2 moles of LiOH
So, [tex]2.44\times 10^{-3}mol[/tex] moles of iron (II) chloride will react with = [tex]\frac{2}{1}\times 2.44\times 10^{-3}=4.88\times 10^{-3}mol[/tex] of LiOH
Calculating the molarity of LiOH by using equation 1:
Moles of LiOH = [tex]4.88\times 10^{-3}mol[/tex]
Volume of LiOH = 35.1 mL = 0.0351 L
Putting values in equation 1, we get:
[tex]\text{Molarity of LiOH}=\frac{4.88\times 10^{-3}mol}{0.0351L}=0.139M[/tex]
Hence, the molarity of LiOH is 0.139 M