$(a)\;89.4g\;cm^{-3}\qquad(b)\;8.94g\;cm^{-3}\qquad(c)\;894g\;cm^{-3}\qquad(d)\;0.894g\;cm^{-3}$

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We know that

$Z = \large\frac{a^3\times d\times N_A}{M}$

$\therefore d = \large\frac{ZM}{a^3N_A}$

Z = 4 for fcc unit cell ;

$M = 63.5 g\; mol^{-1}$

$a = 361.6\times 10^{-10}cm$

$N_A = 6.023\times10^{23}$

Putting these values in (i) we get

$d = \large\frac{4\times63.5}{(361.6\times10^{-10})^3\times 6.023\times10^{23}}$

$=8.94 g\;cm^{-3}$

Hence answer is (B)

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