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.
$$\begin{align}
C_1 \cdot V_1 &amp;= C_2 \cdot V_2 \\
C_1 &amp;= 10~\mathrm{mg/l} \\
V_1 &amp;= 2~\mathrm{ml} \\
V_2 &amp;= (2~\mathrm{ml} + 10~\mathrm{ml})= 12~\mathrm{ml} \\
C_2 &amp;= \frac{C_1 \cdot V_1}{V_2}  \\
&amp;= \frac{(10~\mathrm{mg/l}) \cdot (2~\mathrm{ml})}{12~\mathrm{ml}} \\
&amp;= 1.67~\mathrm{mg/l}
\end{align}$$
Is this correct?
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.