an element has a BCC structure with cell edge of 288pm. the density of the element is 7.2g/cm3. how many atom are present in 208g of the element?
Question: Is this math?
To calculate the number of atoms present in a given mass of an element, you need to use the formula:
Number of atoms = (Mass of the substance / Atomic mass) * Avogadro's constant
Given data:
Cell edge (a) = 288 pm = 288 × 10^(-12) m
Density (ρ) = 7.2 g/cm^3 = 7.2 × 10^3 kg/m^3
Mass (m) = 208 g
To further calculate the number of atoms, we need to determine the volume of the unit cell and the atomic mass.
1. Calculate the volume of the unit cell (V):
Since the element has a body-centered cubic (BCC) structure, the unit cell volume (V) can be calculated as:
V = (a^3) / (4)
= (288 × 10^(-12))^3 / 4
= 2.35505 × 10^(-23) m^3
2. Calculate the atomic mass (M):
The atomic mass needs to be determined based on the element in question. For the sake of example, let's assume the atomic mass is 50 g/mol.
3. Calculate the number of atoms (N):
Using the formula, N = (m / M) * (1 / V) * (1 / Avogadro's constant), where Avogadro's constant is 6.022 × 10^23 mol^-1.
N = (208 / 50) * (1 / 2.35505 × 10^(-23)) * (1 / 6.022 × 10^23)
= 0.17447 * 4.2342 × 10^23
≈ 7.38 × 10^22 atoms
Therefore, there are approximately 7.38 × 10^22 atoms in 208 g of the element.