# Question #9c740

Jul 8, 2017

${\text{55.5 g CaCl}}_{2}$

#### Explanation:

For starters, notice that your sample contains

$3.01 \cdot {10}^{23} = \frac{1}{2} \cdot \textcolor{b l u e}{6.02 \cdot {10}^{23}}$

formula units of calcium chloride. As you know, in order to have $1$ mole of calcium chloride, you need to have a sample that contains $\textcolor{b l u e}{6.02 \cdot {10}^{23}}$ formula units of calcium chloride $\to$ this is known as Avogadro's constant.

In your case, the sample contains exactly half, $\frac{1}{2}$, of a number equal to Avogadro's constant of formula units of calcium chloride, which means that it contains $0.5$ moles of calcium chloride.

To convert the number of moles to grams, you must use the compound's molar mass.

You will end p with

$0.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CaCl"_2))) * "110.98 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(darkgreen)(ul(color(black)("55.5 g}}}}$

The answer is rounded to three sig figs, the number of significant figures you have for the number of formula units present in the sample.