Question

The number of atoms in $100$ g of a fcc crystal with density of $10.0$ g cm${}^{-3}$ and cell edge equal to $200$ pm is equal to :

• A. $5\times {10}^{24}$
• B. $5\times {10}^{25}$
• C. $6\times {10}^{23}$
• D. $2\times {10}^{25}$

Answer 1

Answer: (A)

$5\times {10}^{24}$

density $=\frac{zM}{N_a\times (a\times {10}^{-10}{)}^3}$

$10=\frac{4\times M}{6\times {10}^{23}\times (200\times {10}^{-10}{)}^3}$

$M=12\text{g/mol}$

Number of atoms is 100 g $=\frac{100}{12}\times 6\times {10}^{23}=5\times {10}^{24}$

Hence, option A is correct.