Palladium crystallizes with a face-centered cubic structure. It has a density of 12.0 , a radius of 138 , and a molar mass of 106.42. What is Avogardo's number?
No units, no workee problem.
Density is 12.0 g/cm3, radius is 138 pm, and molar mass is 106.42 g/mol
Worked the same way as the previous post about solving for atomic mass except here you are given the atomic mass and the unknown is the 6.02 x 10^23. The remainder of the problem mirrors the other one.
See the previous post about converting pm to cm first.
5,17
To find Avogadro's number, we need to use the given information about the density, radius, and molar mass of palladium.
Avogadro's number (represented by N๐) is defined as the number of atoms or molecules in one mole of a substance. It is approximately equal to 6.022 ร 10^23 particles per mole.
First, let's calculate the volume of one palladium atom. Since palladium crystallizes with a face-centered cubic (FCC) structure, each unit cell contains 4 atoms. The volume of a single atom can be determined using the formula for the volume of a sphere:
V = (4/3) * ฯ * r^3
where V is the volume and r is the radius.
V = (4/3) * ฯ * (138 pm)^3
V = (4/3) * 3.14 * (138 pm)^3
V โ 2.54 ร 10^-22 cm^3
Next, we need to calculate the volume of one mole of palladium atoms. Since each unit cell contains 4 atoms, the volume of one mole of palladium atoms will be:
V_mole = (4 atoms / unit cell) * V_atom
The density of palladium (represented by ฯ) is given as 12.0 g/cm^3. We can use this information along with the molar mass (M) to calculate the volume of one mole:
V_mole = M / ฯ
V_mole = 106.42 g/mol / 12.0 g/cm^3
V_mole โ 8.87 cm^3/mol
Finally, we can find Avogadro's number (N๐) by dividing the volume of one mole of palladium atoms by the volume of a single atom:
N๐ = V_mole / V_atom
N๐ โ 3.49 ร 10^23 atoms/mol
Therefore, Avogadro's number is approximately 3.49 ร 10^23.