Answer:
[tex]M_2=3.34x10^{-3}M[/tex]
Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
[tex]M_1V_1=M_2V_2[/tex]
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
[tex]M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M[/tex]
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