How do I find a concentration from a sample tested on a spectrophotometer?

Question

I have an assignment to calculate the concentration from a sample. The assignment is:

You have 100 µL of RNA. You take 10 µL from the sample and put it in 990 µL of water in the cuvette. The OD is 0.05. Calculate the concentration of RNA in the sample. Calculate how many µL you have to take to make a solution of 100 ng.

Thus far I have succeeded to calculate the concentration from the OD. It is 2 µg/µL. I used the formula that I found on the internet :

$$OD\times \text{Standard coeff.}(\pu{40 µg/ml})\times \text{Sample dilution}$$ but I cannot figure out how from that I can calculate the solution with 100 ng, because there is not mentioned the final volume for 100 ng, or how to calculate how much I have to take µL from the cuvette. I do not understand where to begin from.

I tried with the formula $C_1V_1=C_2V_2$, but I could not figure that out.

Answer 1

The question being asked is, how many uL of your solution do you need to withdraw in order for that volume to contain 100 ng of the RNA. If there are 2 ug/uL in the RNA solution, that is equivalent to 2000 ngm/uL. So if you want 100 ng of RNA, you will need to withdraw $$\ce{\frac{2000 ngm}{1 uL}~=~ \frac{100 ngm}{x ~uL}}$$ solving for x yields an answer of 1/20 uL or 0.05 uL. 0.05 uL of the RNA solution will contain 100 ng of the RNA.

Answer 2

You make at least one mistake in your original calculation: the dilution factor should be $\times 100$ not $\times 1000$, which importantly reduces the concentration in the original sample to $\pu{0.2 \mu g / \mu L}$ or $\pu{200 ng / \mu L}$ RNA and makes it rather trivial to see what volume is required to obtain $\pu{100 ng}$, clearly this must be $\frac12 \pu{\mu L}$, which is reasonable (as opposed to $\pu{0.05 \mu L}$ which isn't, as noted in comments, because pipetting such a small amount is very difficult). When you obtain an unreasonable result it should make you suspect that you made a mistake somewhere.

Another point is that you can conceptualize what volume of stock contains enough of the substance (encloses the mass desired), such that if the solution is more dilute you need more of the solution, that is $V \propto c^{-1}$. If you want a sample with $\pu{100 ng}$ of RNA you just need an aliquot of the stock such that $c_{stock} \times V_\text{aliquot} = \pu{100 ng}$. This should of course provide the same result as the equation written in the other answer.