Question

The number of atoms in 100 g of a fcc crystal with density $=10.0{\text{g/cm}}^3$ 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

The correct option is A $5\times {10}^{24}$

$\rho =\frac{Z\times M}{a^3\times N_A}$
or $M=\frac{10\times (200{)}^3\times {10}^{-30}\times 6\times {10}^{23}}{4}=12$ $(\because 1\text{pm}={10}^{-10}\text{cm})$

Thus, 12 g contains $=N_A=6\times {10}^{23}\text{atoms}$
$\therefore$ 100 g will contain $=\frac{6\times {10}^{23}}{12}\times 100$
$=5\times {10}^{24}$ atoms