Answer:
[tex]C_2=0.97M[/tex]
Explanation:
Hello,
In this case, for dilution process, we can notice that the initial moles remain the same once the dilution is completed, therefore, both concentration and volume change considering:
[tex]n_1=n_2\\\\V_1C_1=V_2C_2[/tex]
In such a way for the given final volume, the resulting concentration is noticed to be:
[tex]C_2=\frac{V_1C_1}{V_2} =\frac{10mL*2.5M}{25.8mL}\\ \\C_2=0.97M[/tex]
This is supported by the fact that the higher the volume the lower the concentration.
Best regards.
