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?




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.


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