Question
The number of atoms in $$100\,g$$ of an $$fcc$$ crystal with density, $$d = 10\,g/c{m^3}$$ and cell edge equal to $$100\,pm,$$ is equal to
A.
$$1 \times {10^{25}}$$
B.
$$2 \times {10^{25}}$$
C.
$$3 \times {10^{25}}$$
D.
$$4 \times {10^{25}}$$
Answer :
$$4 \times {10^{25}}$$
Solution :
$$\eqalign{
& M = \frac{{\rho \times {a^3} \times {N_A} \times {{10}^{ - 30}}}}{Z} \cr
& = \frac{{10 \times {{\left( {100} \right)}^3} \times 6.02 \times {{10}^{23}} \times {{10}^{ - 30}}}}{4} \cr
& = 15.05 \cr
& \therefore \,\,{\text{Number of atoms in }}100{\text{ }}g \cr
& = \frac{{6.02 \times {{10}^{23}}}}{{15.05}} \times 100 \cr
& = 4 \times {10^{25}} \cr} $$