Calculating the concentration of a diluted solution

I have this problem:

From a stock solution of initial concentration of $$10~\mathrm{mg/l}$$, I take $$2\,\mathrm{ml}$$, and then I add $$10\,\mathrm{ml}$$ of distilled water.

I am supposed to calculate the final concentration. I have solved the problem in this way.

\begin{align} C_1 \cdot V_1 &= C_2 \cdot V_2 \\ C_1 &= 10~\mathrm{mg/l} \\ V_1 &= 2~\mathrm{ml} \\ V_2 &= (2~\mathrm{ml} + 10~\mathrm{ml})= 12~\mathrm{ml} \\ C_2 &= \frac{C_1 \cdot V_1}{V_2} \\ &= \frac{(10~\mathrm{mg/l}) \cdot (2~\mathrm{ml})}{12~\mathrm{ml}} \\ &= 1.67~\mathrm{mg/l} \end{align}

Is this correct?

• Dear anonymous user, please do not replace mg/l with ppm, as they are entirely different concepts and often not even remotely interchangeable! – Jan Sep 23 '15 at 14:53

1 Answer

The answer is basically correct (see note at the end of my answer about significant figures) but there is a simpler method, using a dilution factor.

The original volume of your solution was 2 mL, and the final volume was 12 mL, so the dilution factor is simply $\frac{2}{12}$.

The original concentration times the dilution factor gives the final concentration:

$$\pu{10\frac{mg}{L}}\cdot\frac{2}{12} = \pu{1.7\frac{mg}{L}}$$

Note that only 2 significant figures were given in the problem and your answer reported 3.