Question

Question #c2f50

Answer 1

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 cm`.

Next, work out the volume of the FCC cell: `2.46` x `10^-24 cm^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 cm^3` = `277.935` g/`cm^3`

Rounding up, closest answer is (c).