# Question f1ee8

Jan 1, 2015

I think you are dealing with a volume-to-volume percent concentration ($\text{v/v%}$), which is defined as the volume of the solute divided by the total volume of the solution and multiplied by 100%. This is mathematically described as

"v/v%" = ("volume of solute")/("total volume of solution") * 100%

So, in your case, you have a volume of solute equal to $50 \mu L$ and a volume of solvent, NOT of total solution, equal to $200 \mu L$.

You add the solute to the solvent and get a total volume of solution of

$50$ $\mu$$\text{L} + 200$ $\mu$$\text{L} = 250$ $\mu$ $\text{L}$, which means that your $\text{v/v}$ percent concentration is

"v/v%" = (50muL)/(250muL) * 100%= 20%#

So, essentially, you have dilluted the original $50 \mu L$ sample to 1/5th of its original concentration:

${C}_{\text{initial") * 50 muL = C_("final") * 250 muL => C_("final") = C_("initial}} \cdot \frac{50 \mu L}{250 \mu L}$

${C}_{\text{final") = C_("initial}} / 5$.