Question

Question #33cb2

Answer 1

`2 * 10^(24)"atoms of Ca"`

Explanation:

The first thing to do here is figure out how many formula units of calcium chloride, `"CaCl"_2`, you have in `3` moles of this ionic compound.

To do that, you must use the definition of a mole, which as you know is simply a very, very large collections of atoms. More specifically, in order to have one mole of an ionic compound, you need to have `6.022 * 10^(23)` formula units of said compound.

`color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"f units"color(white)(a/a)|))) ->` Avogadro's number

Now, your sample will contain a total of

`3 color(red)(cancel(color(black)("moles CaCl"_2))) * (6.022 * 10^(23)"f units CaCl"_2)/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = 1.8 * 10^(24)"f units CaCl"_2`

As you know, a formula unit of calcium chloride is made up of

• one atom of calcium, `1 xx "Ca"`
• two atoms of chlorine, `2 xx "Cl"`

Since every formula unit of calcium chloride contains one atom of calcium, it follows that your answer will be

`1.8 * 10^(24) color(red)(cancel(color(black)("f units CaCl"_2))) * "1 atom Ca"/(1color(red)(cancel(color(black)("f unit CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2 * 10^(24)"atoms of Ca")color(white)(a/a)|)))`

The answer is rounded to one sig fig, the number of sig figs you have for the number of moles of calcium chloride.