[tex]1.75\times 10^{-5} M[/tex] was the concentration of the original solution.
Explanation:
[tex]M_1V_1=M_2V_2[/tex]
Where:
[tex]M_1[/tex]=The concentration of the solution before dilution
[tex]V_1[/tex]= Volume of solution before dilution
[tex]M_2[/tex]=The concentration of the solution after dilution
[tex]V_2[/tex]= Volume of solution after dilution
Given:
The 1.00 mL of solution was diluted with 4.00 mL of water.
The concentration of a diluted solution is [tex]3.5\times 10^{-6 M}[/tex].
To find:
The original concentration of the solution before dilution.
Solution:
The original concentration of solution before dilution = [tex]M_1[/tex]
The volume of original solution used = [tex]V_1 = 1.00 mL[/tex]
The volume of water added for dilution = v = 4.00 mL
The concentration of solution after dilution = [tex]M_2=3.5\times 10^{-6 }M[/tex]
The volume of diluted solution used = [tex]V_2[/tex]
[tex]V_2= V_1+v=1.00 mL+4.00 ml=5.00 mL[/tex]
Using the equation of dilution:
[tex]M_1V_1=M_2V_2\\M_1=\frac{M_2V_2}{V_1}\\M_1=\frac{3.5\times 10^{-6 M}\times 5.00 mL}{1.00 mL}\\M_1=1.75\times 10^{-5}M[/tex]
[tex]1.75\times 10^{-5} M[/tex] was the concentration of the original solution.
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