# If copper, density = 9.0 g/cm^3 and atomic mass 63.5, bears face-centered

If copper, density = 9.0 g/cm^3 and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?

A. 0.0028

B. 0.0235

C. 0.0011

D. 0.0323

This interesting question is from Solid State in division Solid State of Chemistry – Class 12

The correct answer is B. 0.0235

Easiest explanation: Density, d of unit cell is given by d = $$\frac{zM}{a^3N_A}$$

Given,

Density, d = 9.0 g/cm^3

Atomic mass, M = 63.5 g/mole

Edge length = a

NA = Avogadro’s number = 6.022 x 10^23

z = 4 atoms/cell

On rearranging the equation for density we get a^3 = $$\frac{zM}{dN_A}$$

Substituting the given values:

a^3 = $$\frac{4 \times 63.5}{9 \times 6.022 \times 10^{23}}$$

Therefore, a = 360.5 pm

The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2$$\sqrt{2}$$r.

On substituting the value of ‘a’ in the given relation, r = $$\frac{360.5}{2\sqrt{2}}$$=127.46 pm

Now, for spherical particles volume, V = 4πr^3/3 and surface area, S = 4πr^2

Required ratio = S/V=4πr^2/(4πr^3/3) = 3/r (after simplifying)

Thus, S/V = 3/127.46 = 0.0235.