At $100^{\circ} C,$ copper $(C u)$ has fcc unit cell structure with cell edge length of $x \circ A$ . what is the approximate density of Cu $(in g c m^{- 3})$ at this temperature?
[Atomic mass of $C u$ $= 63.55 u]$

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Solution

Copper $(C u)$ has fcc unit cell structure, therefore, Z = 4
$d e n s i t y, d = \frac{4 \times M}{N_{A} (a)^{3}}$
$d = \frac{63.5 \times 4}{6 \times 10^{23} \times x^{3} \times 10^{- 24}} g / c m^{3}$

$d = \frac{63.5 \times 4 \times 10}{6 \times x^{3}} g / c m^{3}$

$d = \frac{423.33}{x^{3}} = (\frac{422}{x^{3}})$

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Similar questions

100^{o} C

(C u)

F C C

x ˚ A

C u

^{- 3}

C u = 63.55 u

3.61 A^{o}

C u = 63.5 g m o l^{- 1}

Copper has the fcc crystal structure. Assuming an atomic radius of 130 pm for copper atom (Cu = 63.54):

(1) What is the length of the unit cell of Cu?

(2) What is the volume of the unit cell?

(3) How many atoms belong to the unit cell?

(4) Find the density of Cu?

130

(C u = 63.54)

3.61 \times 10^{- 8} c m

63.5 g m o l^{- 1}

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