# Question 4e23c

Aug 9, 2016

${\text{1.5 mol dm}}^{- 3}$

#### Explanation:

The first thing to do here is to pick a volume of this ammonium sulfate, ("NH"_4)_2"SO"_4, solution and use its density to find its mass.

Since molarity is defined as moles of solute per cubic decimeters of solution, pick a ${\text{1-dm}}^{3}$ sample of this solution to make the calculations easier.

This

1 color(red)(cancel(color(black)("dm"^3))) * (10^3"mL")/(1color(red)(cancel(color(black)("dm"^3)))) = 10^3"cm"^3

sample has a density of ${\text{1.10 g cm}}^{- 3}$, which means that you get a mass of $\text{1.10 g}$ for every cubic centimeter of solution. In your case, the sample will have a mass of

10^3 color(red)(cancel(color(black)("cm"^3))) * "1.10 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "1100 g"

Now, this solution is said to be 18% by mass. This tells you will get $\text{18 g}$ of solute for every $\text{100 g}$ of solution. Your sample will contain

1100color(red)(cancel(color(black)("g solution"))) * ("18 g"color(white)(a)("NH"_4)_2"SO"_4)/(100color(red)(cancel(color(black)("g solution")))) = "198 g"color(white)(a)("NH"_4)_2"SO"_4

To find the number of moles of ammonium sulfate present in $\text{198 g}$, use the compound's molar mass

198 color(red)(cancel(color(black)("g"))) * ("1 mole"color(white)(a)("NH"_4)_2"SO"_4)/(132.14color(red)(cancel(color(black)("g")))) = "1.4984 moles"color(white)(a)("NH"_4)_2"SO"_4

Finally, the molarity of the solution will be -- keep in mind that your sample has a volume of ${\text{1 dm}}^{3}$

c = "1.4984 moles"/"1 dm"^3 = color(green)(|bar(ul(color(white)(a/a)color(black)("1.5 mol dm"^(-3))color(white)(a/a)|)))#

The answer is rounded to two sig figs, the number of sig figs you have for the concentration of the soltuion.