# Question 91322

May 13, 2016

Approximately $\text{330 g}$.

#### Explanation:

The first thing to do here is figure out how many grams of calcium chloride, ${\text{CaCl}}_{2}$, are needed to make your target solution.

As you know, molarity is defined as number of moles of solute per liter of solution. This basically means that a $\text{1 molar}$ solution will contain $1$ mole of solute in one liter of solution.

In your case, you have $3$ liters of $\text{1 molar}$ solution, which means that you have

3 color(red)(cancel(color(black)("L solution"))) * overbrace("1 mole CaCl"_2/(1color(red)(cancel(color(black)("L solution")))))^(color(purple)("= 1 molar")) = "3 moles CaCl"_2#

To convert the number of moles of calcium chloride to grams, use the compound's molar mass

$3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)"330 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the volume and molarity of the target solution.