Question

When gold crystallizes, it forms face-centred cubic cells.The unit cell edge length 408pm. Calculate the density of gold. Molar mass of gold is 197 g/mol.​

Answer 1

Answer:

ANSWER

Atomic mass of gold is 197 g/mol. 1 mole of gold weighs 197 g. 1 mole of gold contains 6.023×10

23

atoms. Hence, 6.023×10

23

atoms of gold weighs 197 g. 1 atom of gold weighs

6.023×10

23

atoms

197 g

A fcc unit cell of gold contains 4 atoms.

The mass of the unit cell of gold = mass of 4 atoms of gold

=

6.023×10

23

atoms

197 g

×4 atoms=1.3×10

−21

g.

Convert the unit from g to Kg.

The mass of the unit cell of gold =1.3×10

−21

g×

1000 g

1 kg

=1.3×10

−24

Kg.

Answer 2

Given :

Edge Length (a) = 408 pm

Molar mass (M) = 197 g/mol

To Find :

Density of Gold

Solution :

Density of crystal (d) = $\frac{n * \frac{A}{N_A} }{a^3}$

n = no. of atoms; A = Atomic mass; N$_A$ = Avogadro's Number; a = edge length.

Density of the gold crystal (d) = $\frac{4 * \frac{197}{6.022 * 10^2^3} }{(408 * 10^{-12} )^3}$

                                      d  = 0.00000193 × 10¹³

                                      d  =  192.67 × 10⁵ gm/m³

Therefore, the density of the gold crystal is found to be 192.67 × 10⁵ gm/m³.