Question

Question #06a7b

Answer 1

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 `"20-mL"` sample of stock solution.

To do that, sue the known composition of the stock solution, which is said to contain `"40 g"` of ammonium nitrate in `"500 mL"` of solution.

Use this as a conversion factor to calculate the number of grams of ammonium nitrate present in `"20 mL"` of stock solution

`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 `"80 mL"` of water to get a 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 `color(blue)(5)`, which means that the stock solution was `color(blue)(5)` times more concentrated than the diluted solution.

Now, the sample of stock solution contained `"1.6 g"` in `"20 mL"` of solution. The diluted solution contains `"1.6 g"` of ammonium nitrate in `"100 mL"` of solution, which means that you have

`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 `"20 mL"` of diluted solution, i.e. the stock solution was `color(blue)(5)` 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 `"1.6 g"` of ammonium nitrate in `"100 mL"` of diluted solution and convert the mass of solute to 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.