Answer:
The final concentration is 0.2497 M.
Explanation:
1. Molarity of the starting solution = [tex]M_1=1.80 M[/tex]
Volume of the starting solution = [tex]V_1=60.0 mL[/tex]
After dilution of the solution to 219 mL.
Molarity of solution after dilution = [tex]M_2[/tex]
Volume of the new solution = [tex]V_2=218 mL[/tex]
[tex]M_1\times V_1=M_2\times V_2[/tex] (Dilution law)
[tex]M_2=\frac{1.80 M\times 60.0 mL}{218 mL}=0.495 M[/tex]
2. Now,109 mL of 0.495 M solution was diluted by 107 mL
Molarity of the solution = [tex]M_1=0.495 M[/tex]
Volume of the solution = [tex]V_1=109 mL[/tex]
Molarity of solution after dilution = [tex]M_2[/tex]
Volume of the new solution(107 mL of water is added) = [tex]V_2=109 mL+107 mL = 216 mL[/tex]
[tex]M_1\times V_1=M_2\times V_2[/tex] (Dilution law)
[tex]M_2=\frac{0.495 M\times 109 mL}{216 mL}=0.2497 M[/tex]
The final concentration is 0.2497 M.
