# Question #03711

##### 1 Answer

#### Explanation:

This one is a little tricky because it wants you to perform a dilution by adding *solute* to the *solvent*, and not the other way around.

Here's how you can approach thus type of problems.

When performing a regular dilution, the **dilution factor** is used to express the ratio between the **concentration** of the stock solution and the **concentration** of the target solution by using the ratio between their two **volumes**.

Starting from the equation for **dilution calculations**

#color(blue)(c_1 xx V_1 = c_2 xx V_2)" "# , where

**stock solution**

**target solution**

you can rearrange the terms to get

#c_1/c_2 = V_2/V_1#

This is the *dilution factor*,

#color(blue)("D.F." = V_2/V_1)#

In your case, you want to perform a

#"D.F." = 38 = V_2/V_1#

Now, **do not** be confused by the fact that you're adding the antibiotic solution to the saline solution!

Let's say that you will end up using a volume of **initial volume** of the solution.

#V_1 = xcolor(white)(a)"mL"#

Since you're adding this solution to **final volume** of the solution will be

#V_2 = "7.91 mL" + xcolor(white)(a)"mL" = (7.91 + x)color(white)(a)"mL"#

Plug this into the equation for the dilution factor to get

#38 = V_2/V_1#

#38 = ((7.91 + x)color(red)(cancel(color(black)("mL"))))/(xcolor(red)(cancel(color(black)("mL")))) = (7.91 + x)/x#

This means that you have

#38 * x = 7.91 + x#

#37 * x = 7.91 implies x = 7.91/37 = 0.214#

Therefore, adding

The answer will thus be

#V_"antibiotic" = color(green)("0.214 mL") -># rounded to thre sig figs

So, remember that dilutions are all about keeping the number of moles of solute (or the amount, if you will) **constant** while **increasing** the volume of the solution.

So always try to think

final volume#-># initial volume#-># dilution factor

or

initial concentration#-># final concentration#-># dilution factor