Question

A metal crystallizes as a body centred cubic lattice with the edge length of unit cell equal to 0.304 nm. If the molar mass of the metal is 50.3 g/mol, its density is-
[A] $5.945\text{ g/c}{{\text{m}}^{3}}$
[B] $2.9725\text{ g/c}{{\text{m}}^{3}}$
[C] $8.9175\text{ g/c}{{\text{m}}^{3}}$
[D] $\text{4}\text{.458 g/c}{{\text{m}}^{3}}$

Answer 1

Hint: We can calculate this by using the formula- density = mass$\div $volume. For mass, we need to calculate the number of atoms in each unit cell in a body centred cubic lattice and put it in the above formula to get the answer.

Complete answer:
BCC lattice is the abbreviation used for body centred cubic lattice.
In a body centred cubic lattice, atoms are arranged in each corner and at the centre of the cube.
To calculate the density, we need to find out the number of atoms in each unit cell of the body centred cubic lattice-
As we know, there are 8 corners in a cube.
Therefore, it gives us a total of 8 atoms in the corners.
Each corner is shared among seven other such unit cells i.e. each corner atom is shared with 8 other unit cells i.e. contribution from each corner atom is $\dfrac{1}{8}$.
Therefore, total contribution of the 8 corner atoms= $8\times \dfrac{1}{8}$= 1
Now, there is one atom in the body centre and it is not shared with any other unit cell. So, its contribution is 1.
Therefore, the total number of atoms in each unit cell = 2.
Now, as we can see it is given in the question, the molar mass of the metal is 50.3 g/mol and it has 2 atoms per unit cell.
Therefore, mass if the metal will be, m = $\dfrac{2\times 50.3}{{{N}_{A}}}=\dfrac{2\times 50.3}{6.023\times {{10}^{23}}}$
The length of the unit cell is given as 0.304 nm
Since, it is a cube its volume will be ${{\left( 0.304\text{ }nm \right)}^{3}}$ or, ${{\left( 0.304\times \text{1}{{\text{0}}^{-7}}cm \right)}^{3}}$
As we know, density = mass$\div $volume,
Putting the value of mass and volume from the above calculation in this formula, we get
Density = $\dfrac{m}{{{a}^{3}}}=\dfrac{\dfrac{2\times 50.3}{6.023\times {{10}^{23}}}}{{{\left( 0.304\times {{10}^{-7}} \right)}^{3}}}=5.945g/c{{m}^{3}}$

Therefore, the correct answer is option [A] .

Note:
We know that a lattice is a network which gives us the three dimensional arrangement of atoms. There are 14 different Bravais lattices that we study in case of solids. The body centred and the cubic close packed structures are also Bravais lattice. BCC is one of the primitive lattices and CCP is just like the face centred Bravais lattice.