A metal crystallizes in such a lattice in which only 70% of the total space of the crystal is occupi

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A metal crystallizes in such a lattice in which only 70% of the total space of the crystal is occupied by the atoms. If the atomic mass of the metal is 32π g/mol and the atomic radius is 0.2 nm, then the density of the metal is

Accepted Answer

(Volume of crystal containing one mole metal)$\times \frac{70}{100} = \left( 6 \times 10^{23} \right) \times \frac{4}{3}\pi \times \left( 0.2 \times 10^{- 7}\text{cm} \right)^{3}$

$\therefore\mspace{6mu}\mspace{6mu} V_{\text{crystal}} = \frac{64}{7}\pi\text{c}\text{m}^{3} \therefore\text{Density = }\frac{32\pi}{\left( \frac{64}{7} \right)\pi} = 3.5\text{gm/c}\text{m}^{3}$

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