The density of cesium is 1870 kg/m3 and the cell volume is 2.23 x 10-22 mL. Determine the number of atoms in the unit cell. Note: the number of atoms in a unit cell is a whole number.
To determine the number of atoms in the unit cell, we need to first calculate its volume in cubic meters (m^3), as the density is given in kg/m^3.
Given:
Density of cesium (ρ) = 1870 kg/m^3
Cell volume (V) = 2.23 x 10^-22 mL
To convert the cell volume to cubic meters, we need to convert milliliters (mL) to cubic meters (m^3). We know that 1 mL is equal to 1 x 10^-6 m^3.
Cell volume in cubic meters (V_m^3) = cell volume in mL x 1 x 10^-6 m^3/mL
= (2.23 x 10^-22 mL) x (1 x 10^-6 m^3/mL)
= 2.23 x 10^-22 x 10^-6 m^3
= 2.23 x 10^-28 m^3
Now that we have the volume of the unit cell, we can use Avogadro's number and the atomic mass of cesium to determine the number of atoms.
Avogadro's number (N_A) = 6.022 x 10^23 atoms/mol
Atomic mass of cesium (molar mass, M) = XX g/mol
To calculate the number of atoms (N) in the unit cell, we can use the formula:
N = (ρ x V) / (M x N_A)
Substituting the given values:
N = (1870 kg/m^3) x (2.23 x 10^-28 m^3) / (XX g/mol x 6.022 x 10^23 atoms/mol)
Note: The atomic mass of cesium is required to complete the calculation. You need to find the molar mass of cesium and substitute the value to obtain the final answer.