# Question f76b6

Apr 23, 2016

$\text{5 mL}$

#### Explanation:

The idea here is that you need to figure out how many milliliters of the stock solution will contain the amount of sucrose that must be present in the target solution, i.e. $\text{0.25 g}$.

In fact, you don't even need to worry about the volume of the target solution in your calculations.

So, you know that your stock solution contains $\text{5 g}$ of sucrose per $\text{100 mL}$ of solution. You can use this concentration as a conversion factor to help you find the volume of the stock solution that would contain $\text{0.25 g}$ of sucrose

$0.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g sucrose"))) * "100 mL solution"/(5color(red)(cancel(color(black)("g sucrose")))) = color(green)(|bar(ul(color(white)(a/a)"5 mL solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that in order to prepare your target solution, you must start with $\text{5 mL}$ of the stock solution and add enough water to make the total volume of the solution equal to $\text{1 L}$, i.e. you must add $\text{995 mL}$ of water.

This is equivalent to diluting the $\text{5-mL}$ sample of the stock solution by a factor of

"D.F." = (1000 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = 200#

The dilution factor is calculated by dividing the volume of the diluted solution by the volume of the stock solution.