Question

gold crystallises in ccp structure. the total number of voids present in 197 g of gold is​

Answer 1

Answer : The total number of voids present in 197 g of gold is, $3N_A$

Explanation : Given,

Mass of gold = 197 g

Number of unit cells in CCP = 4

First we have to calculate the total number of voids.

As we know that,

Voids in Octahedral + Tetrahedral sites = 2n + n = 3n

The total voids = 3n

The number of gold atoms = $\frac{197g}{197g/mole}\times N_A=N_A$

The total number of voids present in gold = $3N_A$

Therefore, the total number of voids present in 197 g of gold is, $3N_A$

Answer 2

Three voids are present in 197 gm of gold.

To find:

The no. of Voids in the 19 grams of gold.

Solution:

Cubic close set packed crystalline structures is also have 74% space. It is similar to the hexagonal structures.

The basic repeating structure in the crystalline solids are refereed to as unit cell. These have tetrahedral and octahedral voids.

Number of units cells in Cubic close set packed crystalline are four .

Number of unit cells =4

Voids in octahedral + tetrahedral sites = 2n+n= 3n

Total voids = 4×3=12

For one atom$\frac{12}{4}$ = 3 voids

Therefore, one gold atom has three voids.