Final concentration of stock solution upon dilution

Question

If we have a stock solution that is $\pu{1 mg/mL}$, and we have $\pu{0.1 mL}$ of that stock to which we add $\pu{0.9 mL}$ of water, what is the final concentration?

For this I assumed I had to use the concentration formula:

$$\mathrm{M_{1}V_{1} = M_{2}V_{2}}$$

I thought the initial concentration was $\pu{1 mg/mL}$, since that is my stock. I thought the initial volume was $\pu{0.1 mL}$, since that is what they needed from the initial stock. I assume the final volume is $\pu{1 mL}$, since that is the total volumes of the components.

$$\mathrm{(1\;mg/mL)\cdot(0.1\;mL) = (M2)\cdot(1\;mL)}$$

$$\mathrm{M_2 = \frac{0.1\;mg}{1\;mL}}$$

$$\mathrm{M_2 = 0.1\;mg/mL}$$

I feel I am right due to the units evaluating properly out, since the original concentration is giving in $\mathrm{mg/mL}$ , the final should be also in $\mathrm{mg/mL}$, right?

Answer 1

You are correct. You might find it more intuitive to think of it this way:

In the end you will have $\pu{1 mL}$ of solution. In that end solution you have $\pu{0.1 mL} \times \frac{\pu{1mg}}{\pu{mL}}$ of stock, or (cancelling the $\pu{mL}$s) $\pu{0.1mg}$ of the chemical in question. Therefore, final concentration is $\frac{\pu{0.1mg}}{\pu{1mL}} = 0.1 \frac{\pu{mg}}{\pu{mL}}$.