Question

An atom crystallises in FCC crystal lattice and has a density of 10g $ {\text{c}}{{\text{m}}^{{\text{ - 3}}}} $ with unit cell edge length of 100pm. Calculate number of atoms present in 1g of crystal.

Answer 1

Hint: FCC is also called as face centered cubic structure. In FCC, atoms are arranged at the corners and center of each cube face. To solve this question, we will use the density formula and equate the two formulae of number of moles.
Formula used: The formulae used in this equation are as follows,
 $ {\text{Density = }}\dfrac{{{\text{Molar mass }} \times {\text{ Z}}}}{{{{\text{N}}_a} \times {\text{ }}{{\text{a}}^3}{\text{ }}}} $
 $ \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}{\text{ = }}\dfrac{{{\text{Number of atoms}}}}{{{{\text{N}}_a}}} $ .

Complete Step by step solution:
The density of the FCC crystal lattice is given as 10g $ {\text{c}}{{\text{m}}^{{\text{ - 3}}}} $ . Let a be the edge length of the unit cell of FCC lattice. The edge length of the unit cell is 100pm. The Avogadro number is represented by $ {N_a} $ . Z is the coordination number and for FCC lattice, the value of Z is 4. Moles can be given by the number of atoms divided by the Avogadro number. It can also be represented as a given mass divided by molar mass.
The given mass of the crystal lattice is 1 g. The number of atoms present in 1 g of crystal lattice is given as follows,
 $ {\text{Density = }}\dfrac{{{\text{Molar mass }} \times {\text{ Z}}}}{{{{\text{N}}_a} \times {\text{ }}{{\text{a}}^3}{\text{ }}}} $ …(i)
 $ \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}{\text{ = }}\dfrac{{{\text{Number of atoms}}}}{{{{\text{N}}_a}}} $
Hence, $ {\text{Molar mass = }}\dfrac{{{\text{Given mass }} \times {\text{ }}{{\text{N}}_a}}}{{{\text{Number of atoms}}}} $ …(ii)
Substituting (ii) in (i),
 $ {\text{Density = }}\dfrac{{{\text{Given mass }} \times {\text{ }}{{\text{N}}_a}{\text{ }} \times {\text{ Z}}}}{{{\text{Number of atoms }} \times {\text{ }}{{\text{N}}_a} \times {\text{ }}{{\text{a}}^3}{\text{ }}}} $
 $ {\text{Number of atoms = }}\dfrac{{{\text{Given mass }} \times {\text{ Z}}}}{{{{\text{a}}^3}{\text{ }} \times {\text{ density}}}} $
 $ {\text{Number of atoms = }}\dfrac{{1 \times 4}}{{{{\left( {100 \times {{10}^{ - 10}}} \right)}^3} \times 10}}{\text{ = }}\dfrac{{4 \times 1}}{{10 \times {{10}^{ - 24}}}}{\text{ = 4}} \times {\text{1}}{{\text{0}}^{23}}{\text{ atoms}} $

Therefore, the number of atoms in 1 g of FCC crystal lattice is $ 4 \times {10^{23}} $ .

Note:
The smallest repeating unit of a crystal structure is known as the unit cell.
Coordination number is the number of the particles that each particle in a crystalline solid contacts. In other words, it is the number of the nearest neighbouring atoms or the ions surrounding an atom or an ion.