How many moles of ammonium nitrate are in 335 mL of 0.425M #NH_4NO_3#?

1 Answer
Mar 9, 2016


#"0.142 moles"#


You could get fancy and say that this solutions contains zero moles of ammonium nitrate, but that would not be the answer the problem is looking for.

Here's why that is.

A solution's molarity will tell you how many moles of solute you get per liter of solution.

The problem with soluble ionic compounds is that they dissociate completely in aqueous solution to form cations and anions. In this case, ammonium nitrate exists in solution as ammonium cations, #"NH"_4^(+)#, and nitrate anions, #"NO"_3^(-)#

#"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)#

This means that your solution contains no ammonium nitrate, so technically the concentration of ammonium nitrate is zero.

Simply put, molarity was initially used as a way to express the concentration of a particular chemical species present in solution. (For more on that, go here).

However, it is now common to use molarity as a was to express the concentration of a solute regardless of the form in which it exists in solutions.

In your case, you know that the solution has a molariy of #"0.425 mol L"^(-1)# and a total volume of #"335 mL"#. To find how many moles of ammonium nitrate you have in solution, use the equation

#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#

Do not forget to convert the volume to liters by using the conversion factor

#"1 L" = 10^3"mL"#

You will have

#c = n/V implies n = c * V#

#n = "0.425 mol" color(red)(cancel(color(black)("L"^(-1)))) * 335 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)(|bar(ul(color(white)(a/a)"0.142 moles"color(white)(a/a)|)))#

The answer is rounded to three sig figs.