# 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 (17.03 g/mol)?

Oct 1, 2016

${\text{10.8 % NH}}_{3}$

#### Explanation:

The idea here is that you need to pick a sample volume of this solution and calculate

• the total mass of this sample
• the mass of ammonia it contains

To make the calculations easier, let's pick a $\text{1000 mL}$ sample. This solution is said to have a density of ${\text{0.950 g mL}}^{- 1}$, which means that every milliliter of solution has a mass of $\text{0.950 g}$.

The sample we've picked will have a mass of

1000 color(red)(cancel(color(black)("mL"))) * "0.950 g"/(1color(red)(cancel(color(black)("mL")))) = "950 g"

Now, you know that this solution has a molarity of $\text{6.00 M}$, which basically means that every liter, which is the equivalent of $\text{1000 mL}$, will contain $6.00$ moles of ammonia.

Since we've picked a sample of $\text{1000 mL}$, you can say that it will contain $6.00$ moles of ammonia. To convert this to grams, use the compound's molar mass

6.00 color(red)(cancel(color(black)("moles NH"_3))) * "17.03 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "102.18 g"

Now, the solution's percent concentration by mass, $\text{%m/m}$, tells you how many grams of ammonia, the solute, you get per $\text{100 g}$ of solution.

You already know how many grams of ammonia you have in $\text{950 g}$ of solution, so use this known composition to figure out how many grams of ammonia you'd get in $\text{100 g}$ of solution

100 color(red)(cancel(color(black)("g solution"))) * "102.18 g NH"_3/(950color(red)(cancel(color(black)("g solution")))) = "10.76 g NH"_3

Therefore, you can say that the solution's mass by mass percent concentration is

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{% m/m" = 10.8%"NH}}_{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.