# Question 9fe6f

Nov 25, 2016

We calculate the mass and the volume of a unit cell, and then we calculate its density to be ${\text{9.23 g/cm}}^{3}$.

#### Explanation:

Calculate the number of atoms in the unit cell.

The unit cell of polonium is a simple cube:

It has a $\text{Po}$ atom at each corner.

But how many $\text{Po}$ atoms are actually inside the cube?

Here's a cutaway diagram of the cube itself.

We see that there is only one-eighth of a Po atom at each corner.

Since there are 8 corners,

$\text{Number of Po atoms" = 8 color(red)(cancel(color(black)("corners"))) × ("⅛ Po atom")/(1 color(red)(cancel(color(black)("corner")))) = "1 Po atom}$

Calculate the mass of the unit cell

"Mass of unit cell" = 1 color(red)(cancel(color(black)("atom Po"))) × (1 color(red)(cancel(color(black)("mol Po"))))/(6.022 × 10^23 color(red)(cancel(color(black)("Po atoms")))) × "209 g"/(1 color(red)(cancel(color(black)("mol Po"))))

= 3.471 × 10^"-22" color(white)(l)"g"

Calculate the volume of the unit cell.

The formula for the volume of a cube is $V = {l}^{3}$, where $l$ is the length of an edge of the cube.

V = l^3 = (0.335 × 10^"-9" color(red)(cancel(color(black)("m"))))^3 × ("100 cm"/(1 color(red)(cancel(color(black)("m")))))^3 = 3.760 × 10^"-23"color(white)(l) "cm"^3

Calculate the density of the unit cell

ρ = m/V = (3.471 × 10^"-22" color(white)(l)"g")/(3.760 × 10^"-23" color(white)(l)"cm"^3) = "9.23 g/cm"^3#

Since density is an intensive property, this is also the density of the bulk metal.