Iron crystallizes in a bcc system with a lattice parameter of $2.861 \dot{\mathrm{A}}$. Calculate the density of iron in the bcc system
[atomic weight of $\mathrm{Fe}=56, \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{mol}^{-1}$].
A. $7.94 \mathrm{g} \mathrm{mL}^{-1}$
B. $8.96 \mathrm{g} \mathrm{mL}^{-1}$
C. $2.78 \mathrm{g} \mathrm{mL}^{-1}$
D. $6.72 \mathrm{g} \mathrm{mL}^{-1}$
Answer
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Hint: We know that the number of atoms per unit cell in a crystal which crystallizes in a bcc system is two. The density of the atom in any of the system varies directly with the number of atoms in the unit cell of the system.
Complete step by step answer:
We can take the value of the molecular mass of the iron which crystallizes in a bcc system from the question which is $56 \mathrm{g} \mathrm{mol}^{-1}$. The given edge length of the unit cell is $2.861 \dot{\mathrm{A}}$.
We know that, $1 \dot{\mathrm{A}}=10^{-8} \mathrm{cm}$. Therefore, we can convert the given value of the lattice parameter in cm. Therefore, we can say that the lattice parameter is equal to $2.861 \times 10^{-8} \mathrm{cm}$.
We know that the formula used for calculating the is given by:
$\rho=\dfrac{Z \times \mathbf{M}}{a^{3} \times \mathbf{N}_{A}}$
Here,{Z is the number of atoms per unit cell,M is the molar mass and a is the lattice parameter.
When we substitute all the values in the above equation we get the value of density as follows.
$\begin{aligned} \rho &=\dfrac{2 \times 56 \mathrm{g} \mathrm{mol}^{-1}}{\left(2.861 \times 10^{-8} \mathrm{cm}\right)^{3} \times 6.02 \times 10^{23} \mathrm{mol}^{-1}} \\ &=7.94 \mathrm{g} \mathrm{mL}^{-1} \end{aligned}$
Thus, we can say that the density of iron in the bcc system is $7.94 \mathrm{g} \mathrm{mL}^{-1}$.
Note:
We know that by knowing the unit-cell dimensions we can calculate the theoretical density of a cubic crystal. The theoretical density obtained using the formula used in the question is with the assumption that each lattice point is occupied by the species. But if some lattice points remain vacant, then we can also calculate the percentage occupancy from the observed and the theoretical densities
Complete step by step answer:
We can take the value of the molecular mass of the iron which crystallizes in a bcc system from the question which is $56 \mathrm{g} \mathrm{mol}^{-1}$. The given edge length of the unit cell is $2.861 \dot{\mathrm{A}}$.
We know that, $1 \dot{\mathrm{A}}=10^{-8} \mathrm{cm}$. Therefore, we can convert the given value of the lattice parameter in cm. Therefore, we can say that the lattice parameter is equal to $2.861 \times 10^{-8} \mathrm{cm}$.
We know that the formula used for calculating the is given by:
$\rho=\dfrac{Z \times \mathbf{M}}{a^{3} \times \mathbf{N}_{A}}$
Here,{Z is the number of atoms per unit cell,M is the molar mass and a is the lattice parameter.
When we substitute all the values in the above equation we get the value of density as follows.
$\begin{aligned} \rho &=\dfrac{2 \times 56 \mathrm{g} \mathrm{mol}^{-1}}{\left(2.861 \times 10^{-8} \mathrm{cm}\right)^{3} \times 6.02 \times 10^{23} \mathrm{mol}^{-1}} \\ &=7.94 \mathrm{g} \mathrm{mL}^{-1} \end{aligned}$
Thus, we can say that the density of iron in the bcc system is $7.94 \mathrm{g} \mathrm{mL}^{-1}$.
Note:
We know that by knowing the unit-cell dimensions we can calculate the theoretical density of a cubic crystal. The theoretical density obtained using the formula used in the question is with the assumption that each lattice point is occupied by the species. But if some lattice points remain vacant, then we can also calculate the percentage occupancy from the observed and the theoretical densities
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