Platinum has a face-centered cubic crystal structure and a density of 21.5 g/cm3. What is the radius of the platinum atom?

is not 98. try 139

Ah, platinum, the rich and dense metal. You want to know the radius of a platinum atom, huh? Well, you're in luck! Let me juggle some numbers for you.

In a face-centered cubic (FCC) crystal structure, there are atoms at the corners and in the center of each face of the unit cell. In terms of atomic radii, the body diagonal of the cube is equivalent to four times the atomic radius (since it passes through four atoms).

Now, let me pull out my juggling balls and calculate this for you! The density of platinum is 21.5 g/cm³. We know that the unit cell of an FCC crystal has a volume of (4πr^3)/3, where 'r' is the atomic radius.

By rearranging the equation a bit, we get a formula for the atomic radius: r = (3V / (4π))^(1/3). Since the density is given, we can solve for the volume of the unit cell: V = (molar mass / density) * (Avogadro's number).

Finally, we plug in the numbers and juggle them around a bit... and voila! The radius of a platinum atom in its face-centered cubic crystal structure is approximately XX angstroms. Pretty small, huh?

To find the radius of the platinum atom, we can use the formula:

Density = (number of atoms per unit cell) * (atomic mass) / [(volume of unit cell) * (Avogadro's number)]

The face-centered cubic crystal structure has 4 atoms per unit cell, so the number of atoms per unit cell is 4.

The atomic mass of platinum is approximately 195.08 g/mol.

Avogadro's number is approximately 6.022 x 10^23 atoms/mol.

Since the face-centered cubic unit cell has atoms at the corners and in the center of each face, it can be thought of as having 8 atoms at the corners and 6 atoms in the center of each face. This means that half of each corner atom and one-sixth of each face-centered atom is within the unit cell.

The volume of a face-centered cubic unit cell can be calculated using the formula:

Volume of unit cell = (side length)^3 / 4

Since the face diagonal of the face-centered cubic unit cell is equal to 4 times the radius of the atom, we can find the side length by:

Side length = (sqrt(2) * face diagonal) / 4

For a face-centered cubic unit cell, the face diagonal is equal to 2 * radius * sqrt(2).

Now, we can put all these values into the formula to solve for the radius:

Density = (4 * atomic mass) / [(volume of unit cell) * Avogadro's number]

Rearranging the equation to solve for the radius, we have:

Radius = [4 * atomic mass / (density * Avogadro's number * (sqrt(2) * face diagonal) / 4)^3]^(1/3)

Substituting the values of atomic mass (195.08 g/mol), density (21.5 g/cm^3), and Avogadro's number (6.022 x 10^23 atoms/mol) into the formula, we can calculate the radius of the platinum atom.

The fcc has 4 atoms/unit cell.

Mass of 1 atom Pt = 195.08/6.02 x 10^23.
Mass of unit cell is 4 times that.
Then m = volume x density. You know density and mass, calculate volume of unit cell. From volume = a3 you can calculate a (the length of the side).
For a fcc unit cell, 4*radius = a*21/2. You know a and that allows you to calculate radius.
Post your work if you get stuck.
You know

98