# Question #2ade7

##### 1 Answer

#### Explanation:

The thing to remember about **dilution calculations** is that the **ratio** that exists between the volume of the *diluted solution* and the volume of the *concentrated solution* is **equal** to the ratio that exists between the concentration of the *concentrated solution* and the concentration of the *diluted solution*.

You thus have

#color(blue)(bar(ul(|color(white)(a/a)color(black)(V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted")|)))#

These two ratios give you the **dilution factor**.

In your case, the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is

#V_"diluted"/V_"concentrated" = (400. color(red)(cancel(color(black)("L"))))/(10. color(red)(cancel(color(black)("L")))) = 40#

This means that the concentration of the *concentrated solution* is **times higher** than the concentration of the *diluted solution*.

#c_"concnetrated"/c_"diluted" = 40#

Therefore, you have

#c_"diluted" = c_"concentrated"/40#

In your case, this will be equal to

#c_"diluted" = ("19.1 mol L"^(-1))/40 = color(darkgreen)(ul(color(black)("0.48 mol L"^(-1))))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the volume of the concentrated solution.