How many grams of ammonium sulfate are needed to make a 0.25 L solution at a concentration of 6 M?

1 Answer
Apr 9, 2016

#"200 g"#


Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate, #("NH"_4)_2"SO"_4#, it must contain.

Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.

So, a solution's molarity tells you how many moles of solute you get per liter of solution.

In this case, the solution must have a molarity of #"6 mol L"^(-1)#, which is equivalent to saying that it must contain #6# moles of ammonium sulfate per liter of solution.

Use this value as a conversion factor to find the number of moles that you'd get in #"0.25 L"# of solution

#0.25 color(red)(cancel(color(black)("L solution"))) * ("6 moles"color(white)(a) ("NH"_4)_2"SO"_4)/(1color(red)(cancel(color(black)("L solution")))) = "1.5 moles" color(white)(a)("NH"_4)_2"SO"_4#

Now use ammonium sulfate's molar mass, which is listed as being equal to #"132.14 g mol"^(-1)#, to find how many grams you'd need in order to get #1.5# moles

#1.5color(red)(cancel(color(black)("moles"color(white)(a)("NH"_4)_2"SO"_4))) * "132.14 g"/(1color(red)(cancel(color(black)("mole"color(white)(a)("NH"_4)_2"SO"_4)))) = "198.21 g"#

Rounded to one significant figure, the answer will be

#"mass of ammonium sulfate" = color(green)(|bar(ul(color(white)(a/a)"200 g"color(white)(a/a)|)))#