What is the final concentration of a solution if 15-mL of a 2.4-M solution was diluted to 100-mL and this process was repeated two more times?

Answer 1

You can use the formula V1M1 = V2M2 where V1 is initial volume; M1 is initial molarity; V2 is final volume; M2 is final molarity.

Looking at the first dilution:

(15 ml)(2.4 M) = (100 ml)(X M) and X = 0.36 M

Second dilution:

(15 ml)(0.36 M) = (100 ml)(X M) and X = 0.054 M

Third dilution?

(15 ml)(0.054 M) = (100 ml)(X M) and X = 0.0081 M

You could also do this another way:

Since each dilution is 15 ml to 100 ml, that is a 100/15 = 6.667 x dilution. Thus, diluting the original three times by this factor (6.67 x 6.67 x 6.67) is = 297 fold dilution. Hence, the final solution will have a molarity of

2.4 M/297 = 0.0081 M