# Question #c2f50

##### 1 Answer
Jun 27, 2016

I'm a bit rusty on these, but the density should be the mass of the 4 atoms of rhodium that make up the face centred cubic cell, divided by the volume of the cell.

The correct answer should be (c).

#### Explanation:

First, convert pm into cm: $1.35$ x ${10}^{-} 8 c m$.

Next, work out the volume of the FCC cell: $2.46$ x ${10}^{-} 24 c {m}^{3}$

Next work out the average mass of 1 atom of rhodium: 102.9 g/mol divided by $6.02$ x ${10}^{23}$ = $1.709$ x ${10}^{-} 22$ g

Next, work out the mass of the 4 rhodium atoms in the FCC unit cell: $1.709$ x ${10}^{-} 22$ g x 4 = $6.837$ x ${10}^{-} 22$ g

Finally, work out the density by dividing this last value by the volume of the FCC cell: $6.837$ x ${10}^{-} 22$ g / $2.46$ x ${10}^{-} 24 c {m}^{3}$ = $277.935$ g/$c {m}^{3}$

Rounding up, closest answer is (c).