A metal having atomic mass 60.22 g/mole crystallizes in ABCABC…. type packing. The density of each m

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A metal having atomic mass 60.22 g/mole crystallizes in ABCABC…. type packing. The density of each metal atom if the edge length of unit cell is 10 Å is (NA = 6.022 × 1023)

Accepted Answer

Packing is FCC for which $\sqrt{2}a = 4r$

=$\therefore r = \frac{\sqrt{2} \times 10\overset{o}{A}}{4} = 2.5\sqrt{2}\overset{o}{A}$

Now, density of metal atom = $\frac{m_{\text{atom}}}{V_{\text{atom}}} = \frac{60.22}{\left( 6.022 \times 10^{23} \right) \times \frac{4}{3}\pi \times \left( 2.5\sqrt{2} \times 10^{- 8} \right)^{3}} = 0.54\text{ gm/c}\text{m}^{3}$

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