An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0 g cm$$^{-3}$$. The number of atoms present in 180 g of the element is _____ $$\times 10^{23}$$. (Nearest integer)

We consider a body-centred cubic (BCC) unit cell with edge length $$a = 300$$ pm $$= 300 \times 10^{-10}$$ cm $$= 3 \times 10^{-8}$$ cm and density $$\rho = 6.0$$ g cm$$^{-3}$$. For BCC, the number of atoms per unit cell is $$Z = 2$$.

Using the density relation $$\rho = \frac{Z \times M}{N_A \times a^3}$$ and solving for $$M$$ gives $$M = \frac{\rho \times N_A \times a^3}{Z}$$. Substituting the known values yields $$M = \frac{6.0 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{2}$$.

Since $$a^3 = (3 \times 10^{-8})^3 = 27 \times 10^{-24} = 2.7 \times 10^{-23}$$ cm$$^3$$, substituting this into the expression for $$M$$ gives $$M = \frac{6.0 \times 6.022 \times 10^{23} \times 2.7 \times 10^{-23}}{2} = \frac{6.0 \times 6.022 \times 2.7}{2} = \frac{97.56}{2} = 48.78$$ g/mol.

To find the number of atoms in 180 g of the element, the number of moles is calculated as $$\text{Number of moles} = \frac{180}{48.78} = 3.69$$ mol. Therefore, the number of atoms is $$\text{Number of atoms} = 3.69 \times 6.022 \times 10^{23} = 22.22 \times 10^{23}$$.

Therefore, the number of atoms present in 180 g is $$22 \times 10^{23}$$ (nearest integer).