Question

The density of KBr is 2.75 $gc{{m}^{-3}}$ . The length of the unit cell is 654 pm. Atomic mass in amu of K = 39, Br = 80. Then, the solid is:
A. Face-centered cubic
B. Simple cubic system
C. Body-centered cubic system
D. None of the above

Answer 1

Hint:The relation between density, unit length and atomic mass of a molecule is as follows.
\[d=\frac{z\times m}{V\times {{N}_{o}}}=\frac{z\times m}{{{a}^{3}}\times {{N}_{o}}}\]
Here d = density of the molecule
M = Molecular mass of the molecule
a = length of the unit cell
${{N}_{o}}$ = Avogadro number

Complete step-by-step answer:- In the question it is given to find the shape of the solid by using the various parameters in the question.
- The density of the KBr is 2.75 $gc{{m}^{-3}}$ , the length of the unit cell = 654 pm.
- By using the below formula we can calculate the number of unit cells present in the KBr structure.
\[d=\frac{z\times m}{{{a}^{3}}\times {{N}_{o}}}\]
Here, Here d = density of the molecule = 2.75 $gc{{m}^{-3}}$
M = Molecular mass of the molecule = 119
a = length of the unit cell = 654$\times {{10}^{-10}}m$
${{N}_{o}}$ = Avogadro number = $6.6023\times {{10}^{23}}$
- Substitute all the known values in the above formula to get the number of unit cells present.
\[ d=\frac{z\times m}{{{a}^{3}}\times {{N}_{o}}} \\
\Rightarrow z=\frac{d\times {{a}^{3}}\times {{N}_{o}}}{m} \\
\Rightarrow z=\frac{2.75\times {{(654\times {{10}^{-10}})}^{3}}\times 6.023\times {{10}^{23}}}{119} \\
\therefore z =4 \]
- Therefore the unit cell contains four molecules of KBr.
- So, the shape of the molecule is a face-centered cubic unit cell.

So, the correct option is A.

Note:We should know the values of length of the unit cell, total molecular weight of the particular chemical and density of the chemical to find the number of unit cells present in the given molecule.