Advertisements
Advertisements
प्रश्न
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Advertisements
उत्तर
Suppose, the edge length of the unit cell = a
Number of atoms present per unit cell = Z
Atomic mass of the element = M
For a cubic unit cell,
Volume of the unit cell = a3 ...(i)
Density of the unit cell
ρ' = `"Mass of unit cell"/"Volume of unit cell"` ...(ii)
Mass of unit cell = Number of atoms per unit cell × Mass of one atom
= Z × m ...(iii)
where m is the mass of a single atom. This is given by
`m = "Atomic mass"/"Avogadro’s number"`
`m = M/N_A` ...(iv)
Substituting the value of m in eq. (iii), we have
Mass of unit cell = `Z xx M/N_A`
Substituting the corresponding values in eq. (ii), the density of the unit cell (ρ') is given by
ρ' = `"Mass of unit cell"/"Volume of unit cell"`
ρ' = `(Z xx M/N_A)/a^3`
or, ρ' = `(Z xx M)/(a^3 xx N_A)`
The density of a crystal (ρ) is the same as the density of its unit cell (ρ'), i.e., ρ' = ρ. Therefore the density ρ of a crystal is given by
`rho = (Z xx M)/(a^3 xx N_A)`
∴ `M = (rho xx a^3 xx N_A)/(Z)`
संबंधित प्रश्न
How many atoms constitute one unit cell of a face-centered cubic crystal?
Gold occurs as face centred cube and has a density of 19.30 kg dm-3. Calculate atomic radius of gold. (Molar mass of Au = 197)
Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.
An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?
An atom located at the body center of a cubic unit cell is shared by ____________.
Volume of unit cell occupied in face-centered cubic arrangement is ____________.
An element (atomic mass 100 g/mol) having bcc structure has unit cell edge 400 pm. The density of element is (No. of atoms in bcc, Z = 2).
The empty space in the body-centered cubic lattice is ____________.
The number of atoms contained in a fcc unit cell of a monoatomic substance is ____________.
Which of the following metal(s) show(s) hexagonal close-packed structure (hcp) and which show face-centered cubic (fcc) structure?
The edge length of fcc cell is 508 pm. If the radius of cation is 110 pm, the radius of anion is:
The density of a metal which crystallises in bcc lattice with unit cell edge length 300 pm and molar mass 50 g mol−1 will be:
An element with atomic mass 100 has a bcc structure and edge length 400 pm. The density of element is:
The percentage of empty space in a body centred cubic arrangement is ______.
The coordination number for body center cubic (BCC) system is
A solid is formed by 2 elements P and Q. The element Q forms cubic close packing and atoms of P occupy one-third of tetrahedral voids. The formula of the compound is ______.
An element A (Atomic weight = 100) having bcc structure has a unit cell edge length 400 pm. The number of atoms in 10 g of A is ______ × 1022 unit cells.
