Question

Or a metal that has the body-centered cubic crystal structure, calculate the atomic radius if the metal has a density of 7.25 g/cm3 and an atomic weight of 50.99 g/mol.

Answer 1

The density of a unit cell lattice is calculated using the following formula:

$\rho =\frac{mn}{a^{3}N_{A}}$...... (1)

Here, $\rho$ is density, m is molar mass or atomic weight, n is number of atoms per unit cell, a is edge length and $N_{A}$ is Avogadro's number.

For a body centered cubic lattice, number of atoms per unit cell is 2 and relation between edge length a and radius of atom r is as follows:

$r=\frac{\sqrt{3}a}{4}$ ...... (2)

Here, r is atomic radius and a is edge length.

Calculate edge length by rearranging equation (1) as follows:

$a=\sqrt[3]{\frac{mn}{\rho N_{A}}}$

Density of metal is $7.25 g/cm^{3}$ and molar mass is 50.99 g/mol, putting the values,

$a=\sqrt[3]{\frac{(50.99 g/mol)(2)}{(7.25 g/cm^{3})(6.023\times 10^{23} mol^{-1})}}=2.86\times 10^{-23} cm$

Therefore, edge length is $2.86\times 10^{-23} cm$, putting the value in equation (2) to calculate atomic radius.

$r=\frac{\sqrt{3}a}{4}=\frac{\sqrt{3}(2.86\times 10^{-8} cm)}{4}=1.24\times 10^{-8} cm$

Therefore, atomic radius will be $1.24\times 10^{-8} cm$.

Answer 2

The atomic radius if the metal has a density of 7.25 g/cm3 and an atomic weight of 50.99 g/mol is 0.124 nm

Further explanation

Or a metal that has the body-centered cubic crystal structure, calculate the atomic radius if the metal has a density of $7.25 g/cm^3$and an atomic weight of 50.99 g/mol.

The density of a metal may be calculated using the following equation:

$\rho = \frac{mn}{a^3 N_A}$

Where:

$\rho$ is density, $m$ is molar mass, $n$ is atomic number per unit cell, $a$ is edge length and $N_A$ is Avogadro's number

Now, for the body-centered cubic crystal structure there are two atoms associated with each unit cell (i.e., $n= 2$), and the atomic radius and unit cell edge length are related as

$r = \frac{\sqrt{3}a }{4}$

Where:

$r$ is atomic radius and $a$ is edge length

Since the unit cell for the body-centered cubic crystal structure has cubic symmetry, $V_C= a^3$. Substitution of the last two relationships into the first equation leads to

$a = \sqrt[3]{\frac{mn}{\rho N_A} }$

$a = \sqrt[3]{\frac{50.99 * 2}{7.25 *6.023 * 10^{23}} }= 2.86 * 10^{-23} cm$

and solving for $R$ yields

$r = \frac{\sqrt{3}a }{4} = \frac{\sqrt{3}* 2.86*10^{-8} }{4}$

$r = 1.24 * 10^-8cm = 0.124 nm$

Learn more

1. Learn more about the body-centered cubic brainly.com/question/4501234
2. Learn more about  the atomic radius brainly.com/question/2631938
3. Learn more about Metal Density brainly.com/question/2284124

Answer details

Grade:  9

Subject:  chemistry

Chapter:  Crystal structure

Keywords: the body-centered cubic,  the atomic radius, Metal Density, an atomic weight, crystal structure