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# An element crystalline in a body centered cubic lattice has an edge of 500 pm. If its density is $4 g cm^{-3}$, the atomic mass of the element (in $g mol^{-1}$) is:(Consider $N_A=6x10^23$)

Last updated date: 18th Sep 2024
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Hint: In body centered cubic lattice (BCC), particles occupy each corner of the unit cell as well as the center of the unit cell. The particle at the center of the unit cell is shared by eight other neighboring unit cells while the particles present at the body center belong to only the particular unit cell it is present in. We can easily calculate the atomic mass of the element by using formula: $\rho =\dfrac{ZM}{{{a}^{3}}\times {{N}_{A}}}$

1: At first, we need to calculate the number of atoms per unit cell of BCC lattice. For the corners, one atom is shared by 8 corners.
2: So, for 8 atoms, total combination from the edge = $\dfrac{1}{8}\times 8=1$ atom.
And one particle in the center remains unshared.
So, the total number of atoms per unit cell = 2. So, Z=2.
3: Now, we calculate the atomic mass of the element using a simple formula:
$\rho =\dfrac{ZM}{{{a}^{3}}\times {{N}_{A}}}$
Where, ρ = density of element
Z = total number of atoms per unit cell
M = mass of the atom
V = volume of the unit cell
4: We need to convert the value of a into cm.
So, a = 500 pm
$= 500x10^{-10}cm$
Now, placing the values in the formula we get,
\begin{align} & 4=\dfrac{2\times M}{{{(500\times {{10}^{-10}})}^{3}}\times 6\times {{10}^{23}}} \\ & \Rightarrow M=\dfrac{4\times 125\times {{10}^{6}}\times {{10}^{-30}}\times 6\times {{10}^{23}}}{2} \\ \end{align}
Solving this, we get:
$M=\dfrac{300\times {{10}^{-1}}}{2}=150g mol^{-1}$
5: Thus, the atomic mass of the element is $150 g mol^{-1}.$

Note:
Students must remember to convert the unit, so that they can cancel out to get the answer. The calculation must be carried out stepwise to avoid mistakes. All the formulae and equations should be kept in handy and memorized by the students.