Question

An element $X$ (At. wt. = 80g/mol) having fcc structure, calculate no. of unit cell in $8gm$ of $X$.

• A. $0.4\times N_A$
• B. $0.1\times N_A$
• C. $\frac{N_A}{40}$
• D. none of these

Answer 1

Answer: (C)

$\frac{N_A}{40}$

$Fcc$ unit cell structure has $4$ atoms in one unit cell,

Let's find the total atoms, $\frac{N}{N_a}=\frac{Wt}{Mwt}$ or $N=\frac{N_a\times 8}{80}=\frac{Na}{10}$

Now as the $Fcc$ unit cell has 4 atoms so

No. of unit cells $=\frac{Na}{10\times 4}=\frac{N_a}{40}$