# Question #73929

##### 1 Answer

#### Explanation:

Your strategy here will be to find the **mass** of the sphere first, then use copper's *density* to find its **volume**.

So, the problem provides you with the number of *atoms* of copper present in the sphere. To find the mass of the sphere, convert the number of atoms to **moles** by using **Avogadro's number**

#1.33 * 10^(24) color(red)(cancel(color(black)("atoms Cu"))) * "1 mole Cu"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Cu"))))#

# = " 2.209 moles Cu"#

Next, use the **molar mass** of copper to find the number of *grams* that would contain that many moles of copper

#2.209 color(red)(cancel(color(black)("moles Cu"))) * "63.546 g"/(1color(red)(cancel(color(black)("mole Cu")))) = "140.4 g Cu"#

Now, you know that copper has a density of

In your case, the sphere will have a volume of

#140.4 color(red)(cancel(color(black)("g Cu"))) * "1 cm"^3/(8.96color(red)(cancel(color(black)("g Cu")))) = "15.67 cm"^3#

All you have to do now is use the formula for the volume of the sphere to calculate its radius.

Rearrange to solve for

#V = 4/3 * pi * r^3 implies r^3 = 3/(4 * pi) * V#

This will get you

#r = root(3) (3/(4pi) * V)#

Plug in your value to find

#r = root(3)(3/(4 * pi) * "15.67 cm"^3) = color(green)(bar(ul(|color(white)(a/a)color(black)("1.55 cm")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.