# Question #06a7b

##### 1 Answer

Here's what I got.

#### Explanation:

The first thing to do here is to figure how many *grams* of ammonium nitrate would be present in the

To do that, sue the known composition of the stock solution, which is said to contain

Use this as a **conversion factor** to calculate the number of grams of ammonium nitrate present in

#20 color(red)(cancel(color(black)("mL stock solution"))) * ("40 g NH"_ 4"NO"_ 3)/(500color(red)(cancel(color(black)("mL stock solution")))) = "1.6 g NH"_ 4"NO"_3#

Now, you add this sample to **diluted solution** that has a volume of

#V_"diluted" = "20 mL" + "80 mL" = "100 mL"#

As you know, the **dilution factor** tells you how **concentrated** the stock solution was compared with the diluted solution by looking at the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

#"DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock"#

In your case, you have

#"DF" = (100 color(red)(cancel(color(black)("mL"))))/(20color(red)(cancel(color(black)("mL")))) = color(blue)(5)#

The dilution factor is equal to **times more concentrated** than the diluted solution.

Now, the sample of stock solution contained

#20 color(red)(cancel(color(black)("mL diluted solution"))) * ("1.6 g NH"_ 4 "NO"_ 3)/(100color(red)(cancel(color(black)("mL diluted solution")))) = "0.32 g NH"_ 4"NO"_ 3#

in **times more concentrated**

#"1.6 g solute / 20 mL solution"/color(blue)(5) = "0.32 g solute / 20 mL solution"#

To convert this to **molarity**, start from the fact that you have **moles** by using the molar mass of ammonium nitrate

#1.6 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_ 4"NO"_ 3)/(80.04color(red)(cancel(color(black)("g")))) = "0.01999 moles NH"_ 4"NO"_ 3#

This will give you a **molarity** of -- *do not* forget to convert the volume to **liters**

#c_"diluted" = ("0.01999 moles NH"_ 4"NO"_ 3)/(100 * 10^(-3)"L") = color(darkgreen)(ul(color(black)("0.2 mol L"^(-1))))#

The answer is rounded to one **significant figure**.