Question

Question #4e23c

Answer 1

`"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 `"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 `"1.10 g cm"^(-3)`, which means that you get a mass of `"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 `"18 g"` of solute for every `"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 `"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 `"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.