Question

An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm3, what is the nature of cubic unit cell? (fcc or ccp)​

Answer 1

Explanation:

Edge length, a=4.05×10

−8

cm=4.05×10

−10

m

Density, d=2.7gcm

−3

=2.7×10

−3

kgm

3

We know that

d=

a

3

×N

A

Z×M

Where, Z is the number of atoms in the unit cell and N

A

is the Avogadro number.

Thus,

Z=

M

d×a

3

×N

A

Z=

0.027

2.7×10

3

(4.05×10

−10

)

3

×6.022×

23

(N

A

=6.022×10

23

)

Z=4orz≈4(fcc)

Since, the number of atoms in the unit cell is 4, the given cubic unit cell has face-centred cubic (fcc) or cubic-close packed (ccp) structure.

Answer 2

Given:

Molar mass of the given element, M=27gmol  −1 =0.027kgmol  −1

Edge length, a=4.05×10  −8  cm=4.05×10  −10  m

Density, d=2.7gcm  −3  =2.7×10  −3  kg m 3

To Find:

The nature of cubic unit cell? (fcc or ccp)

Explanation:

 We know that  

d=  a  3  ×NA  Z×M

Where, Z is the number of atoms in the unit cell and NA ​  is the Avogadro number.  

Thus,

Z=    $\frac{d\times a^{3} \times NA}{M}$            

Z=$\frac{ 7\times10^{3} (4.05\times10 −10 ) 3\times6.022\times23}{0.027}$  

(N  A  =6.022×$10^{23}$  )

Z=4 or z≈4 (fcc)

Since, the number of atoms in the unit cell is 4, the given cubic unit cell has face-centred cubic (fcc) or cubic-close packed (ccp) structure.