# Question #89d0d

##### 1 Answer

#### Answer:

#### Explanation:

For starters, you accidentally added the density of *titanium*, but I assume that the question is about a *copper* sphere. The density of copper is

#rho_"Cu" = "8.96 cm"^3#

The first thing to do here is to find the **mass** of the sphere by using the number of atoms it contains.

To do that, start by converting the number of atoms of copper to **moles** by using **Avogadro's constant**. As you know, in order to have **one mole** of an element, you need to have **atoms** of that element.

In your case, the sphere will contain

#1.14 * 10^(24) color(red)(cancel(color(black)("atoms Cu"))) * overbrace("1 mole Cu"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Cu")))))^(color(blue)("Avogadro's constant"))#

# = " 1.893 moles Cu"#

Next, use the **molar mass** of copper to convert the number of moles to **grams**. As you know, copper has a molar mass of

#1.893 color(red)(cancel(color(black)("moles Cu"))) * "65.546 g"/(1color(red)(cancel(color(black)("mole Cu")))) = "124.1 g"#

Now, in order to find the **volume** of the sphere, you must use its **density**. You will have

#124.1 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(8.96color(red)(cancel(color(black)("g")))) = "13.85 cm"^3#

All you have to do now to find the radius of the sphere is use the formula that gives you the volume of a sphere

#color(purple)(bar(ul(|color(white)(a/a)color(black)(V = 4/3 * pi * r^3)color(white)(a/a)|)))#

Rearrange to solve for

#3 * V = 4 * pi * r^3#

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

Plug in your values to find

#r = root(3)(3/(4pi) * "13.85 cm"^3) = color(green)(bar(ul(|color(white)(a/a)color(black)("1.49 cm")color(white)(a/a)|)))#

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