# Final concentration calculation

I have this problem. From a stock solution of initial concentration of $10\,\mathrm{\frac{mg}{l}}$ I take $2\,\mathrm{ml}$, and then I add $10\,\mathrm{ml}$ of distilled water. I want to calculate the final concentration. I have solved the problem in this way.

$$C_1 \times V_1 = C_2 \times V_2 \\ C_1 = 10\,\mathrm{\frac{mg}{l}} \\ V_1 = 2\,\mathrm{ml} \\ V_2 = (2\,\mathrm{ml} + 10\,\mathrm{ml})= 12\,\mathrm{ml} \\ C_2 = ~?\\ C_2 = \frac{C_1 \times V_1}{V_2} = \frac{10\,\mathrm{\frac{mg}{l}} \times 2\,\mathrm{ml}}{ 12\,\mathrm{ml}} = 1.67 \,\mathrm{\frac{mg}{l}}$$

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

The original volume of your solution was 2 mL, and the final volume was 12 mL, so the dilution factor is simply $\frac{2}{12}$.
$$\pu{10\frac{mg}{L}}\cdot\frac{2}{12} = \pu{1.7\frac{mg}{L}}$$