# If 2.0 mu"L" of a sample must be diluted by a factor of 50, what volume of solvent should be added to accomplish this?

Dec 29, 2017

Mix $2.0$ µ"L" of sample with a bit less than $98$ µ"L" of solvent, then add solvent until you reach the $100$ µ"L" mark.

#### Explanation:

You can use the dilution formula:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {V}_{1} {c}_{1} = {V}_{2} {c}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this formula to get

V_1 = V_2 × c_2/c_1

V_1 = ?; color(white)(mmll)" "" "c_1 = c_1

${V}_{2} = \text{100 µL"; " "" } {c}_{2} = {c}_{1} / 50$

${V}_{1} = \text{100 µL" × (stackrelcolor(blue)(1)color(red)(cancel(color(black)(c_1)))//50)/color(red)(cancel(color(black)(c_1))) = "100 µL" × 1/50 = "2 µL}$

∴ Add a bit less than $98$ µ"L" of solvent to $2$ µ"L" of the sample, then fill up to the mark.

We don't quite add $98$ µ"L" right away because we want to account for any solution volume changes (up/down) due to mixing.