Question

Determine the density of caesium chloride which crystallizes in a bcc type structure with the edge length 412.1 p.m The atomic masses of Cs and Cl are 133 and 35.5 respectively.

Answer 1

Here only one $CsCl$ molecule is present in one bcc unit cell.
To solve this problem we use the formulae,
$\text{Density of crystal = }\dfrac{\text{Mass of unit cell of crystal}}{\text{Volume of unit cell of the crystal}}$

Complete step by step answer:
- In the question it is asked to determine the density of caesium chloride if its edge length of the unit cell is 412.1pm and it is given that the unit cell which generates the entire caesium chloride crystal is bcc.
- We should first know what unit cell means and about the bcc type of unit cell.
- The unit cell is the basic or say the small building block of the crystal which generates the entire crystal structure. In the bcc (body centered cubic unit cell) one type of atom will be placed at the corner and one atom placed in the middle of the unit cell which is not shared by other atoms. The eight atoms placed in the cubic lattice is shared with eight other atoms hence its altogether the contribution is 1 and therefore there are two atoms present in the bcc lattice.
- Here in $CsCl$,$C{{s}^{+}}$ is placed on the middle of the unit cell and the $C{{l}^{-}}$ ions in the corners of the cell.
- To find the density of the crystal we should have the value for the molecular mass of $CsCl$
Add up the atomic mass of Cs and Cl, which gives the molecular mass of $CsCl$.
The atomic mass of Cs and Cl is 133 and 35.5 respectively.
$\text{Molecular mass=133+35}\text{.5=168}\text{.5g/mol}$
- To find the density we have to find the mass of the unit cell and the volume of the unit cell/
The mass of the unit cell is calculated by dividing the molar mass of the $CsCl$ with Avogadro number.
$\text{Mass of unit cell =}\dfrac{\text{Molar mass of CsCl}}{\text{Avogadro number}}$
The value of Avogadro number is $6.022\times {{10}^{23}}$
$\text{Mass of unit cell =}\dfrac{168.5}{6.022\times {{10}^{23}}}=2.8\times {{10}^{-22}}g$
Now find the volume of the unit cell, since the value of edge length is given, take the cube of the given value to get the volume of the unit cell.
$\text{Volume of unit cell=}{{\text{a}}^{3}}$
$\text{Volume of unit cell=(412}\text{.1}\times \text{1}{{\text{0}}^{-10}}{{\text{)}}^{3}}=6.99\times {{10}^{-23}}\text{c}{{\text{m}}^{3}}$
$\text{Density of crystal=}\dfrac{\text{Mass of unit cell of crystal}}{\text{Volume of unit cell of the crystal}}$
$\text{Density of crystal=}\dfrac{\text{2}\text{.8 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-22}}}}{\text{6}\text{.99 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-23}}}}\text{=4}\text{.0057g/c}{{\text{m}}^{\text{3}}}$

Note: Caution should be taken while calculating the exponential powers raised to base 10.
- Alternative method:
We could find the answer by solving the problem directly using the density equation, which is,
$\text{density( }\\!\\!\rho\\!\\!\text{ )=}\dfrac{\text{Z }\\!\\!\times\\!\\!\text{ M}}{{{\text{N}}_{\text{0}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{a}}^{\text{3}}}}$
Z=1
- M is the molecular mass=$\text{168}\text{.5g/mol}$
${{\text{N}}_{\text{0}}}\text{=Avogadro number=6}\text{.022 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}$
$\text{Volume of unit cell=}{{\text{a}}^{3}}=6.99\times {{10}^{-23}}c{{m}^{3}}$
- Substitute all these values in the density equation:
$\text{density( }\\!\\!\rho\\!\\!\text{ )=}\dfrac{\text{1 }\\!\\!\times\\!\\!\text{ 168}\text{.5}}{\text{6}\text{.022 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ }\\!\\!\times\\!\\!\text{ 6}\text{.99 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-23}}}}\text{=4}\text{.0029g/c}{{\text{m}}^{\text{3}}}$