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
#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
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
#"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 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.