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Answer 1 · 2015-01-01T10:35:07

I think you are dealing with a volume-to-volume percent concentration ( $v/v%$ $"v/v%"$ ), 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

$v/v% = \frac{volume of solute}{total volume of solution} \cdot 100 %$ $"v/v%" = ("volume of solute")/("total volume of solution") * 100%$

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

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

$50$ $50$ $μ$ $mu$ $L + 200$ $"L" + 200$ $μ$ $mu$ $L = 250$ $"L" = 250$ $μ$ $mu$ $L$ $"L"$ , which means that your $v/v$ $"v/v"$ percent concentration is

$v/v% = \frac{50 μ L}{250 μ L} \cdot 100 % = 20 %$ $"v/v%" = (50muL)/(250muL) * 100%= 20%$

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

$C_{initial} \cdot 50 μ L = C_{final} \cdot 250 μ L \Rightarrow C_{final} = C_{initial} \cdot \frac{50 μ L}{250 μ L}$ $C_("initial") * 50 muL = C_("final") * 250 muL => C_("final") = C_("initial") * (50muL)/(250muL)$

$C_{final} = \frac{C_{initial}}{5}$ $C_("final") = C_("initial")/5$ .

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