A 51.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 208 mL. A 104 mL portion of that solution is diluted by adding 169 mL of water. What is the final concentration?Assume the volumes are additive.

In this problem the dilution has occurred two times. So the symbol initial concentration, concentration after first dilution and the concentration after second dilution is given as C1, C2, and C3 respectively. Same can be done for the volume i.e, V1, V2, and V3 will be the initial volume, the volume taken after first dilution, and the volume after second dilution.

So, let's use the dilution formula for the first dilution

[tex]C_1 \times V_1 = C_2 \times V_2[/tex]

[tex]C_2 = \frac{C_1\times V_1}{V_2}\\\\C_2 = \frac{1.50\times 51.0}{208}\\\\C_2 = 0.368 M[/tex]

Now, for the second dilution

[tex]C_2 \times V_2 = C_3 \times V_3\\\\C_3 = \frac {C_2 \times V_2}{V_3}\\\\C_3 = \frac {0.368 \times 104}{169}\\\\C_3 = 0.226 M[/tex]