Question

A substance with FCC lattice has density 6250 kg/m3

and Molecular weight 60.2, calculate lattice

constant.​

Answer 1

Explanation:

Given d=8.966 g/cm

Molar mass of copper =63.5 g/mol

N

A

=6.022×10

23

d=

V×N

A

Z×M

where Z=4 (For FCC)

V=

d×N

A

Z×M

V=

8.966×6.023×10

23

4×63.5

V=4.704×10

−23

cm

3

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Answer 2

Answer:

The lattice constant 'a' of a substance with FCC lattice is equal to 4×10⁻¹⁰m.

Explanation:

The formula for density, ρ $=\frac{Z*M}{N_{A}*a^{3}}$                                         ..............(1)

Given, the density of unit cell (ρ) $=6250kgm^{-3} =6.25gcm^{-3}$

The molecular weight, M = 60.2gmol⁻¹

Number of lattice points/atoms in FCC lattice, Z = 4

Avogadro number $N_{A}=6.023*10^{23}mol^{-1}$

Here, 'a' is the edge length of unit cell or lattice constant.

Put the value of ρ, Z, M and $N_{A}$ in equation (1);

$6.25=\frac{(4)*(60.2)}{(6.023*10^{23}*a^{3})}$

$a^{3}=64*10^{-24}$

$a=4*10^{-8}cm$

$a=4*10^{-10}m$

Therefore, the lattice constant of the substance will be equal to 4×10⁻¹⁰m.