If iridium has a density of 23.3 g/cm3 and forms a face-centered cubic lattice, what is the atomic radius of
the iridium atom?
a. 135.7 pm
b. 203.7 pm
c. 271.4 pm
d. 648.0 pm
135.7
which one is correct?
To determine the atomic radius of iridium, we need to consider its density and crystal structure.
1. First, let's recall that face-centered cubic (FCC) lattice has atoms located at the corners and centers of each face of the unit cell. This means that there are 4 atoms per unit cell.
2. The formula to calculate the volume of a unit cell in a face-centered cubic lattice is given as:
Volume of unit cell = (4 * atomic radius^3) / 3
3. We are given the density of iridium as 23.3 g/cm^3. Since we know the density and the molar mass of iridium (we can assume it to be 192.22 g/mol), we can calculate the number of moles in a unit cell using the formula:
Number of moles = density / molar mass
4. The volume of a unit cell can also be calculated using the formula:
Volume of unit cell = (molar mass / Avogadro's number) * Number of moles
5. Equating the two expressions for the volume of the unit cell, we can solve for the atomic radius:
(4 * atomic radius^3) / 3 = (molar mass / Avogadro's number) * Number of moles
6. Rearranging the equation, we find:
Atomic radius = [(3 * molar mass * Number of moles) / (4 * Avogadro's number)]^(1/3)
7. Plugging in the given values, we have:
Atomic radius = [(3 * 192.22 g/mol * density / molar mass) / (4 * Avogadro's number)]^(1/3)
8. Substituting the given density value of 23.3 g/cm^3 and the value of Avogadro's number (6.022 x 10^23 mol^-1), we can calculate the atomic radius.
Calculating the atomic radius using the formula above, we find that the atomic radius of iridium is approximately 135.7 pm. Therefore, the correct option is a. 135.7 pm.