Question

The element chromium exists as a bcc lattice whose unit cell edge is . The density of chromium is $ 7.20\;gc{m^{ - 3}} $ . How many atoms does $ 52.0\;g $ of chromium contain?
(A) $ 3.1 \times {10^{23}}\,atoms/mol $
(B) $ 6.05 \times {10^{23}}\,atoms/mol $
(C) $ 1.2 \times {10^{24}}\,atoms/mol $
(D) $ 12 \times {10^{24}}\,atoms/mol $

Answer 1

Hint: Chromium is a transition metal of d-block with an atomic number of $ 24 $ . It is a steely-grey hard and lustrous element. The atoms of chromium are arranged in the BCC arrangement in space. Chromium has many anti-corrosive properties.
Formula Used:
The density of an atom in a BCC lattice is given as:
 $ D = \dfrac{{z \times M\,}}{{{a^3} \times {N_A}}} $
Where, $ z = $ Number of atoms in that particular lattice (for bcc it is $ 2 $ )
 $ M = $ mass of the atom, $ a = $ edge length of the atom.

Complete step by step solution:
Every crystal lattice is formed from a unit cell. A unit cell is the basic entity of a lattice. A unit cell is the smallest repeating unit of the cell which when repeated over and over gives the crystal lattice. A lattice constant or parameter is used to define the physical dimension of unit cells in a crystal lattice. Generally, there are six lattice parameters of a unit cell, three are edges and three are the angles between them. The edges are termed as $ a,b\,and\,c $ and the angles are $ \alpha ,\beta \,and\,\gamma $ .
In a Body-centred cubic unit cell, atoms are placed at each corner and the centre of the cell. In BCC there are a total $ 8 $ corners and $ 1 $ centre. So the total Number of atoms in an FCC structure is $ 2 $ . Now, we have to calculate the number of atoms in the cell with the help of density. The density in a BCC lattice is given by the formula;
 $ D = \dfrac{{z \times M\,}}{{{a^3} \times {N_A}}} $
We are given, $ z = 2 $ , $ M = 52 $ , $ D = 7.20 $ and $ a = 2.88 \times {10^{ - 10}} $ . Putting these values in the formula we get;
 $ 7.20 = \dfrac{{2 \times 52\,}}{{{{(2.88 \times {{10}^{ - 10}})}^3} \times {N_A}}} $
 $ {N_A} = \dfrac{{2 \times 52 \times {{10}^{ - 3}}\,}}{{{{(2.88 \times {{10}^{ - 10}})}^3} \times 7.20}} = 6.05 \times {10^{23}} $
 $ {N_A} = 6.05 \times {10^{23}}\,atoms/mole $
Hence, $ $ 52.0\;g $ $ of chromium will contain $ 6.05 \times {10^{23}}\,atoms/mol $ .

Therefore, option (B) is correct.

Note:
Unit cells are divided into two categories: (a) Primitive unit cell (B) Centred Unit cell. In the primitive unit cell, the constituent atom or particle occupies the positions of the corners. Centred Unit cells are divided into three types:
-Body Centred Unit cell
-Face Centred Unit cell
-Edge Centred Unit cell