# Question 55f68

Jun 30, 2016

$\text{D.F.} = 20$

#### Explanation:

It's worth mentioning that dissolving your $\text{60 mg}$ sample of solute in $\text{50 mL}$ of solvent will produce a solution of unknown concentration.

That is the case because you don't know the identity of the solute, and consequently can't use its molar mass to figure out how many moles you have in that sample.

However, the concentration of the initial solution is not important here.

The important thing to keep in mind here is that the $\text{5 mL}$ aliquot you take from this $\text{50 mL}$ solution will have the same concentration as the $\text{50 mL}$ solution.

You then dilute the aliquot to a total volume of $\text{100 mL}$. The trick here is to realize that increasing the volume of the solution by a factor will cause its concentration to decrease by the same factor.

This factor is called the dilution factor

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{D.F." = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the dilution factor is equal to

"D.F." = (100 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = 20#

This means that the concentration of the resulting solution, i.e. the diluted solution, will be $20$ times lower than that of the aliquot

${c}_{\text{diluted" = 1/20 * c_"concentrated}}$