# How much (in "L") of a "1.2-M" "NaCl" solution must be used to dilute to create "1 L" of a "0.12-M" "NaCl" solution?

Apr 8, 2018

$\text{0.1 L}$

#### Explanation:

The thing to remember about dilution calculations is that the dilution factor, $\text{DF}$, can be calculated by taking

1. The ratio that exists between the concentration of the stock solution and the concentration of the diluted solution
2. The ratio that exists between the volume of the diluted solution and the volume of the stock solution

So for any dilution, you have

"DF" = color(white)(overbrace(color(black)(c_"stock"/c_"diluted"))^(color(blue)("concentration ratio: stock/diluted"))) = color(white)(overbrace(color(black)(V_"diluted"/V_"stock"))^(color(blue)("volume ratio: diluted/stock")))

In your case, the dilution factor is equal to

"DF" = (1.2 color(red)(cancel(color(black)("M"))))/(0.12color(red)(cancel(color(black)("M")))) = color(blue)(10)

This tells you that the volume of the diluted solution is $\textcolor{b l u e}{10}$ times the volume of the stock solution.

You can thus say that you have

${V}_{\text{stock" = V_"diluted"/"DF}}$

which, in your case, will get you

V_"stock" = "1 L"/color(blue)(10) = color(darkgreen)(ul(color(black)("0.1 L")))

The answer is rounded to one significant figure.

So, in order to perform this dilution, you take $\text{0.1 L}$ of a $\text{1.2-M}$ stock solution of sodium chloride and add enough water to get the final volume of the solution to $\text{1 L}$.

This will decrease the concentration of sodium chloride from $\text{1.2 M}$ in the stock solution to $\text{0.12 M}$ in the diluted solution.