Question

A molecule, ${A_2}B$ (molecular weight = $166.4g/mol$) occupies triclinic lattice with $a = 5{A^o},b = 8{A^o}$ and $c = 4{A^o}$. If the density of ${A_2}B$ is $5.2gc{m^{ - 3}}$, the number of molecules present in one unit cell is__________.
A. $2$
B. $3$
C. $4$
D. $5$

Answer 1

Hint:A crystal lattice is made up of a very large number of unit cells in which every lattice point is occupied by a constituent atom. A single unit cell can be visualized as a three-dimensional structure with one or more atoms. Thus, the number of atoms present in a unit cell can be found by relating the density, molecular mass, and volume of a single unit cell.
Formula Used:
$\rho = \dfrac{{Z \times M}}{{abc \times {N_A}}}$
Where $\rho $ is the density, $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $a,b,c$ represents the unit cell dimensions and ${N_A}$ is the Avogadro number.

Complete step by step solution:
Since the density of a material is constant throughout every unit cell, we can say that the density of a unit cell will be the product of the mass of the unit cell divided by its volume. Mass of the unit cell in turn will be equal to the mass of a single atom multiplied by the number of atoms per unit cell. And volume of the unit cell is given by multiplying its dimensions. Therefore, we have the following relation:
$\rho = \dfrac{{Z \times M}}{{abc \times {N_A}}}$
Where $\rho $ is the density, $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $a,b,c$ represents the unit cell dimensions and ${N_A}$ is the Avogadro number.
From the data given, we have: $\rho = 5.2gc{m^{ - 3}}$, $M = 166.4g/mol$ and ${N_A} = 6.022 \times {10^{23}}$
Since density is given in terms of centimetres, we should convert the unit cell dimensions into that unit too. Therefore,
$a = 5{A^o} = 5 \times {10^{ - 8}}cm$. Similarly, $b = 8 \times {10^{ - 8}}cm$ and $c = 4 \times {10^{ - 8}}cm$. Substituting these values, we get:
$5.2 = \dfrac{{Z \times 166.4}}{{5 \times {{10}^{ - 8}} \times 8 \times {{10}^{ - 8}} \times 4 \times {{10}^{ - 8}} \times 6.022 \times {{10}^{23}}}}$
Rearranging this, we get:
$ \Rightarrow \dfrac{{5.2 \times 5 \times {{10}^{ - 8}} \times 8 \times {{10}^{ - 8}} \times 4 \times {{10}^{ - 8}} \times 6.022 \times {{10}^{23}}}}{{166.4}} = Z$
On solving this, we get:
$Z = 3$
Hence, the correct option to be marked is B.

Note:
-The number of molecules present per unit cell varies with the type of unit cell, and is dependent on the spatial arrangement of each molecule inside the lattice.
-Properties of crystalline solids depend on the nature of interactions between their constituent particles. Crystalline solids are thus characterized by sharp melting points, anisotropic nature and characteristic shapes.