Question #c2f50

1 Answer
Simon Moore
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 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).

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