# Question #4393b

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to

pick a sample of this ammonia solutionuse the solution'sdensityto find themassof the sampleuse themolar massof ammonia to find themass of solute

To make the calculations easier, pick a **molarity** is defined as *moles of solute* per **liter of solution**.

In this case, **moles** of ammonia.

Now, you know that this solution has a density of **mass** of

#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 **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**, **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

#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* *of solution*, you can say that the solution's

#"% 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**.