Question #4393b

Your strategy here will be to

• pick a sample of this ammonia solution
• use the solution's density to find the mass of the sample
• use the molar mass of ammonia to find the mass of solute

To make the calculations easier, pick a #"1 L"# sample of solution. As you know, molarity is defined as moles of solute per liter of solution.

In this case, #"1 L"# of #"6.00 M"# ammonia solution will contain #6.00# moles of ammonia.

Now, you know that this solution has a density of #"0.950 g mL"^(-1)#. Use it to find the mass of #"1 L"# of solution

#1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("0.950 g"/(1color(red)(cancel(color(black)("mL")))))^(color(blue)("the given density")) = "950 g"#

You know that this sample contains #6.00# moles of ammonia, your solute. Use its molar mass to convert this to grams of solute

#6.00 color(red)(cancel(color(black)("moles NH"_3))) * overbrace("17.04 g"/(1color(red)(cancel(color(black)("mole NH"_3)))))^(color(purple)("the given molar mass")) = "102.24 g"#

Now, a solution's mass by mass percent concentration, #"% m/m"#, tells you how many grams of solute you have in #"100 g"# of solution.

#color(blue)(|bar(ul(color(white)(a/a)"% m/m" = "grams of solute / 100 g solution"color(white)(a/a)|)))#

In this case, you know that #"950 g"# of solution contain #"102.24 g"# of solute, which means that #"100 g"# of solution will contain

#100 color(red)(cancel(color(black)("g solution"))) * ("102.24 g NH"_3)/(950color(red)(cancel(color(black)("g solution")))) = "10.8 g NH"_3#

Since this is how many grams of ammonia you get per #"100 g"# of solution, you can say that the solution's #"%m/m"# will be equal to

#"% m/m" = color(green)(|bar(ul(color(white)(a/a)color(black)(10.8%)color(white)(a/a)|)))#

The answer is rounded to three sig figs.