Question

The number of atoms in 100 g of a fcc crystal with density 10 g cm${}^{-3}$ and cell edge as 200 pm is equal to

• A. $3\times {10}^{24}$
• B. $5\times {10}^{24}$
• C. $1\times {10}^{24}$
• D. $2\times {10}^{24}$

Answer 1

Answer: (B)

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

As we know, $M=\frac{\rho \times a^3\times N_A}{n}$ $=\frac{10\times (200\times {10}^{-10}{)}^3\times (6.02\times {10}^{23})}{4}=12.04$
Number of atoms in $100$ g $=\frac{6.02\times {10}^{23}}{12.04}\times 100=0.5\times {10}^{25}$