First slide

Number of atoms per unit cell

Question

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is $2.72 g {cm}^{- 3}$ The molar mass of the metal is $(N_{A} Avogadro's constant = 6.02 \times 10^{23} {mol}^{- 1})$

Moderate

A

$27 g {mol}^{- 1}$
B

$20 g {mol}^{- 1}$
C

$40 g {mol}^{- 1}$
D

$30 g {mol}^{- 1}$

Solution

$\begin{array}{l} d = \frac{ZM}{N_{A} a^{3}} \\ M = \frac{d \times N_{A} \times a^{3}}{Z} \\ = \frac{2 .72 \times 6 .023 \times 10^{23} \times {(404 \times 10^{- 10})}^{3}}{4} \\ M = 27 g {mol}^{- 1} \end{array}$

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