# An ammonia solution is 6.00 M and the density is 0.950 g/mL. What is the mass/mass percent concentration of “NH”_3 (17.04 g/mol)? Thank you

##10.8%##

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, is defined as moles of 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 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 , ##”% 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 .

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