# Question 89d0d

Oct 14, 2016

$\text{1.49 cm}$

#### 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

${\rho}_{\text{Cu" = "8.96 cm}}^{3}$

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 $6.022 \cdot {10}^{23}$ 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"))

$= \text{ 1.893 moles Cu}$

Next, use the molar mass of copper to convert the number of moles to grams. As you know, copper has a molar mass of ${\text{63.546 g mol}}^{- 1}$, which means that the sphere will have a 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

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{V = \frac{4}{3} \cdot \pi \cdot {r}^{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Rearrange to solve for $r$

$3 \cdot V = 4 \cdot \pi \cdot {r}^{3}$

$r = \sqrt{\frac{3}{4 \pi} \cdot V}$

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.