The element Sc has hcp packing with a hexagonal unit cell. The density of scandium is 3.00E3 kg/m3 and the cell volume is 5.00E-26 L. Calculate the value of Avogadro's number to 3 significant figures based on these data. Note: the value may differ from the listed value.
I could convert 3.00E3 kg/m^3 to density of ?? g/cc.
Then mass of unit cell is volume x density = xx grams.
The hexagonal close packing produces 2 atoms/unit cell; therefore, that is the mass for 2 atoms. Divide by 2 to obtain the mass of 1 atom.
Then look up the mass of Sc from the periodic table and that mass is the mass of a mole which contains Avogadro's number of atoms.
Then,
N = atomic mass x (1 atom/?? g) = xx atoms. I worked through this very fast and obtained about 6 x 10^23.
Check my thinking. Check my work.
My answer that i got is 5.99E22
i followed your step all the way till the periodic table..
from there i got Sc atomic mass as 44.96 g
then i multiply it by i atom/7.50E-22
then i got the answer 5.99E22....is my steps and answer correct?
yes. That's the answer I obtaied also. On the others you posted above, I didn't go through any of them in detail. They follow the same general principles. But if you have trouble, post your work and tell us in detail about what you don't understand.
To calculate the value of Avogadro's number based on the given data, we need to use the formula:
Avogadro's number = (density * cell volume) / (atomic mass * molar volume)
Given:
Density of scandium (ρ) = 3.00E3 kg/m^3
Cell volume (V) = 5.00E-26 L
We need to convert the cell volume from liters to cubic meters to match the units of density. 1 liter is equal to 1E-3 cubic meters, so the cell volume can be written as:
V = 5.00E-26 L * (1E-3 m^3 / 1 L) = 5.00E-29 m^3
The atomic mass of scandium (m) is 44.956 grams/mole or 0.044956 kg/mole.
The molar volume (V_m) for a hexagonal close-packed (hcp) structure is given by:
V_m = (a^2 * h * sqrt(3)) / 2
where a is the length of an edge and h is the height.
For an hcp lattice, the relationship between a and h is:
h = (2 * sqrt(6) / 3) * a
or
a = (3 / (2 * sqrt(6))) * h
Since we have the unit cell volume (V), we can substitute the values in the equation to find the molar volume.
V_m = V / number of atoms in the unit cell
In an hcp structure, there are 2 atoms per unit cell.
Therefore, V_m = V / 2
Substituting V = 5.00E-29 m^3:
V_m = 5.00E-29 m^3 / 2 = 2.50E-29 m^3
Now we can calculate Avogadro's number:
Avogadro's number = (density * cell volume) / (atomic mass * molar volume)
= (3.00E3 kg/m^3 * 5.00E-29 m^3) / (0.044956 kg/mole * 2.50E-29 m^3)
Simplifying this equation, we get:
Avogadro's number = (3.00E3 * 5.00E-29) / (0.044956 * 2.50E-29)
Evaluating this expression gives:
Avogadro's number ≈ 6.67E23
Therefore, the value of Avogadro's number based on the given data is approximately 6.67E23 to 3 significant figures.