A copper complex crystallising in a CCP lattice with a cell edge of 0.4518 nm has been revealed by employing X-ray diffraction studies. The density of a copper complex is found to be 7.62 g cm$$^{-3}$$. The molar mass of copper complex is ___ gmol$$^{-1}$$
(Nearest integer): [Given : N$$_A$$ = 6.022 $$\times 10^{23}$$ mol$$^{-1}$$]

In a CCP (cubic close-packed, i.e., FCC) lattice, the number of formula units per unit cell is $$Z = 4$$. The cell edge is $$a = 0.4518 \text{ nm} = 0.4518 \times 10^{-7} \text{ cm} = 4.518 \times 10^{-8} \text{ cm}$$.

The density formula for a crystal is $$\rho = \frac{Z \times M}{N_A \times a^3}$$, where $$M$$ is the molar mass. Rearranging: $$M = \frac{\rho \times N_A \times a^3}{Z}$$.

Computing $$a^3$$: $$(4.518 \times 10^{-8})^3 = 4.518^3 \times 10^{-24}$$. Now $$4.518^3 = 4.518 \times 4.518 = 20.412$$, and $$20.412 \times 4.518 = 92.22$$. So $$a^3 = 92.22 \times 10^{-24} \text{ cm}^3 = 9.222 \times 10^{-23} \text{ cm}^3$$.

Substituting: $$M = \frac{7.62 \times 6.022 \times 10^{23} \times 9.222 \times 10^{-23}}{4} = \frac{7.62 \times 6.022 \times 9.222}{4}$$.

Evaluating the numerator: $$7.62 \times 6.022 = 45.89$$, and $$45.89 \times 9.222 = 423.2$$. Dividing by 4: $$M = \frac{423.2}{4} = 105.8 \approx 106 \text{ g/mol}$$.

The answer is $$106$$.