a metal crystallizes in the face centered cubic crystal structure with a unit cell edge of 5.48 x10 power of -8 cm. the density of the metal is 6.90 g per cubic cm.

a. what is the mass (g) of a single atom of this element.
b. what is the atomic weight of the element (g per mol).
c. what is the radius, in cm, of an atom of the element..

thank you..

edge^3 = volume

mass = volume x density
fcc = 4atoms/unit cell so mass/4 = mass 1 atom
1 atom x 6.02E23 = atomic mass.
4*radius = edge*21/2</sup
Post your work if you get stuck.

can u please explain to me where did u get 2 to find radius?..

To answer these questions, we need to use some basic concepts of crystal structures, unit cells, and atomic weights.

a. To find the mass of a single atom, we need to know the volume of the unit cell and the density of the metal. Since we are given the edge length of the unit cell, we need to calculate the volume of the unit cell.

The volume of a face-centered cubic (FCC) unit cell can be calculated using the formula V = (a^3)/4, where "a" is the length of the edge of the unit cell.

V = (5.48 x 10^(-8) cm)^3 / 4
V ≈ 7.715 x 10^(-24) cm^3

Now, we can calculate the mass of the unit cell by multiplying the volume of the unit cell by the density of the metal.

Mass = Volume x Density
Mass = 7.715 x 10^(-24) cm^3 x 6.90 g/cm^3
Mass ≈ 5.32735 x 10^(-23) g

Since there is only one atom in a unit cell of FCC crystal, the mass of a single atom is the same as the mass of the unit cell.

Therefore, the mass of a single atom of this element is approximately 5.32735 x 10^(-23) grams.

b. To find the atomic weight of the element, we need to know the mass of one mole of atoms. The atomic weight is usually expressed in grams per mole (g/mol).

We know the mass of one atom from the previous calculation. Now, we need to find the Avogadro's number, which is the number of atoms in one mole.

Avogadro's number (NA) is 6.022 x 10^23 atoms/mol.

To calculate the atomic weight, we divide the mass of one atom by Avogadro's number:

Atomic weight (g/mol) = Mass of one atom / Avogadro's number
Atomic weight (g/mol) ≈ 5.32735 x 10^(-23)g / 6.022 x 10^(23)
Atomic weight (g/mol) ≈ 0.088461 g/mol

Therefore, the atomic weight of the element is approximately 0.088461 grams per mole.

c. To find the radius of an atom of the element, we can use some relationships between the unit cell edge length and the atomic radius in an FCC crystal.

In an FCC crystal, the relationship between the unit cell edge length (a) and the atomic radius (r) is:

a = 4√(2) * r

We are given the edge length of the unit cell, so we can rearrange the equation to solve for the atomic radius (r):

r = a / (4√(2))

Substituting the given value for the unit cell edge length:

r = (5.48 x 10^(-8) cm) / (4√(2))
r ≈ 3.8724 x 10^(-8) cm

Therefore, the radius of an atom of this element is approximately 3.8724 x 10^(-8) cm.