Question

A crystalline solid of pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is \[8\;g\;c{m^3}\] then the number of atoms present in 256 g of the crystal is N/times \[{10^{24}}\]. The value of N is:

Answer 1

Hint: The smallest symmetrical repeating unit that will unite repeatedly to form the big symmetrical structure is the unit cell. Each unit cell has a specific lattice point. Lattice point is the interaction of two or more lines that will give the maximum possibility where the atom is placed in the unit cell. The question asked can be solved easily by using the density formula.

Complete Step by step answer: As we know, Atoms per unit cell can be known by knowing that atoms are placed at the centre and at the corner and sharing of the atoms by other similar unit cells which unite to form the big crystal.
Mass of unit cell = number of atoms in unit cell × mass of each atom = z × m
Where, z = number of atoms in a unit cell.
m = Mass of each atom.
Mass of an atom can be given with the help of Avogadro number and molar mass as: \[\dfrac{M}{{\;{N_A}}}\]
Where, M = molar mass.
\[{N_A}\]= Avogadro’s number.
Volume of unit cell, \[V{\rm{ }} = {\rm{ }}{a^3}\]
\[ = > {\rm{ }}Density{\rm{ }}of{\rm{ }}unit{\rm{ }}cell{\rm{ }} = \;\dfrac{{mass\;of\;unit\;cell}}{{volume\;of\;unit\;cell}}\]
\[ = > {\rm{ }}Density{\rm{ }}of{\rm{ }}unit{\rm{ }}cell{\rm{ }} = \;\dfrac{m}{V}\; = \;\dfrac{{z\; \times \;m}}{{{a^3}}}\; = \;\dfrac{{z\; \times \;m}}{{{a^3} \times {N_A}}}\;\]
Given: The edge length is:
\[a = 400\;pm\; = \;400\; \times \;{10^{ - 10}}cm\]
The density formula is \[D = \dfrac{{ZM}}{{{N_0}{a^3}}}\]
Substitute the given values in the above formula.
\[
8\;g/c{m^3} = \dfrac{{4\;atoms \times M}}{{6.023\; \times {{10}^{23}}atoms/mol \times {{(400 \times \;{{10}^{ - 10}}cm)}^3}}}\\
The\;atomic\;mass\;M = \;77g/mol
\]
The number of atoms in 256g of the pure substance \[ = \dfrac{{256g}}{{77g/mol}} \times 6.023\; \times \;{10^{23}} = \;2 \times \;{10^{24}}\]
Which is equal to \[N \times \;{10^{24}}\]
\[\therefore \,N = 2\]

Hence, we conclude that the value of N is two.

Note: PARTICLES PER UNIT CELL FOR SIMPLE, BCC AND FCC IS GIVEN AS:
The number of particles per unit cell is called unit cell constant(z). The general expression for z is\[z = \dfrac{{{N_c}}}{8} + \dfrac{{{N_f}}}{2} + \dfrac{{{N_e}}}{4} + \dfrac{{{N_i}}}{1}\], where c, f, e, i represent corner, face centre, edge centre and inside respectively.