# What volume of 3.0 M #HNO_3 (aq)# must be diluted to make 4.0 L of 0.20 M #HNO_3(aq)#?

##### 1 Answer

#### Explanation:

The key here is the ratio that exists between the **initial concentration** of the solution, i.e. the concentration of the *concentrated solution*, and the **final concentration** of the solution, i.e. the concentration of the *diluted solution*.

This ratio

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

is called the **dilution factor** and it tells you how concentrated the initial solution was compared with the diluted solution. In this case, you have

#"DF" = (3.0 color(red)(cancel(color(black)("M"))))/(0.20color(red)(cancel(color(black)("M")))) = 15#

This means that the concentrated solution was diluted by a factor of **dilution** is that the number of moles of solute **remains constant**.

You can thus **decrease** the concentration of a given solution by **increasing** its volume while keeping the number of moles of solute *unchanged*.

Now, the only way to have a diluted solution that is **times less concentrated** than the concentrated solution is for it to have a *volume* that is **times bigger** than the volume of the concentrated solution.

The dilution factor can thus be expressed in terms of the volume of the concentrated solution,

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

You will have

#V_"concentrated" = V_"diluted"/"DF"#

Plug in your values to find

#V_"concentrated" = "4.0 L"/15 = "0.26667 L"#

Rounded to two **sig figs** and expressed in *milliliters*, the answer will be

#V_"concentrated" = color(green)(|bar(ul(color(white)(a/a)color(black)("270 mL")color(white)(a/a)|)))#

So, to make