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  • Cu form FCC crystal lattice having a = x Ao. The density (g/cm3 ) of its solid is, if M.M. = 63.5 g/mol. Jee main 2019

Answers (1)

best_answer

As we know, 

the density = mass / volume 

in FCC unit cell Z = 4 

So mass = 63.5 x 4

So,

{d=\frac{63.5 \times 4}{6 \times 10^{23} \times x^{3} \times 10^{-24}} \mathrm{g} / \mathrm{cm}^{3}}

{d=\frac{63.5 \times 4 \times 10}{6} \mathrm{g} / \mathrm{cm}^{3}}

{d=\frac{423.33}{x^{3}} \simeq\left(\frac{422}{x^{3}}\right)}.

Hence the correct option is (1).

Posted by

Pankaj Sanodiya

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