# Question 22cf9

Nov 16, 2015

$\text{0.017 M}$

#### Explanation:

Think about what you have here - you know that the original volume of the solution was $\text{100 mL}$. This volume included the $\text{0.23 g}$ of calcium sulfate.

As you know, molarity is defined as moles of solute, which in your case is calcium sulfate, divided by liters of solution.

To get the number of moles of calcium sulfate, use its molar mass. A compound's molar mass tells you what the mass of one mole of that compound is.

In this case, calcium sulfate has a molar mass of $\text{136.14 g/mol}$, which means that every mole of calcium sulfate has a mass of $\text{136.14 g}$.

This means that the $\text{0.23 g}$ of calcium sulfate will contain

0.23color(red)(cancel(color(black)("g"))) * "1 mole CaSO"_4/(136.14color(red)(cancel(color(black)("g")))) = "0.001690 moles CaSO"_4

You now know the number of moles of solute and the volume of the solution, which means that its molarity will be - do not forget to convert the volume to liters!

$c = \frac{n}{V}$

c = "0.001690 moles"/(100 * 10^(-3)"L") = color(green)("0.017 M")#

I left the answer rounded to two sig figs.

SIDE NOTE At room temperature, calcium sulfate has a solubility of about 0.21 g per 100 mL of water, so your solution is very, very close to being saturated.

Of course, that would depend on its temperature, but assuming that you're at ${25}^{\circ} \text{C}$, you can say that your solution is close to saturation.