Question #c2f50

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Question #c2f50

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Answer 1 · 2016-06-27T15:41:19

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: $1.35$ $1.35$ x $10^{- 8} c m$ $10^-8 cm$ .

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

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

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

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

Rounding up, closest answer is (c).

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