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.

Here, [tex]\rho[/tex] is density, m is molar mass or atomic weight, n is number of atoms per unit cell, a is edge length and [tex]N_{A}[/tex] 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:

[tex]r=\frac{\sqrt{3}a}{4}[/tex] ...... (2)

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

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

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

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

[tex]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[/tex]

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

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

Therefore, atomic radius will be [tex]1.24\times 10^{-8} cm[/tex].

tifanihayyu tifanihayyu · Answer 2 · 2019-09-22T13:46:43+02:00

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 [tex]7.25 g/cm^3[/tex]and an atomic weight of 50.99 g/mol.

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

[tex]\rho = \frac{mn}{a^3 N_A}[/tex]

Where:

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

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

[tex]r = \frac{\sqrt{3}a }{4}[/tex]

Where:

[tex]r[/tex] is atomic radius and [tex]a[/tex] is edge length

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

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

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

and solving for [tex]R[/tex] yields

[tex]r = \frac{\sqrt{3}a }{4} = \frac{\sqrt{3}* 2.86*10^{-8} }{4}[/tex]

[tex]r = 1.24 * 10^-8cm = 0.124 nm[/tex]

Learn more

Learn more about the body-centered cubic https://brainly.com/question/4501234
Learn more about the atomic radius https://brainly.com/question/2631938
Learn more about Metal Density https://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