Advertisements
Advertisements
प्रश्न
Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = 27 g mol–1)
Advertisements
उत्तर
Given,
Mass of Al = 8.1
Atomic mass of Al = 27 g mol-1
Number of atoms = number of moles × 6.022 × 1023 = `8.1/27` × 6.022×1023
= 0.3 × 6.022 × 1023
= 1.8066×1023
As one fcc unit cell have 4 atoms
Hence number of unit cells having 1.8066×1023 atoms = `(1.8066 xx 10^23)/4 = 4.51 xx 10^22` unit cell
APPEARS IN
संबंधित प्रश्न
An element with molar mass 27 g mol−1 forms a cubic unit cell with edge length 4.05 ✕ 10−8 cm. If its density is 2.7 g cm−3, what is the nature of the cubic unit cell?
Distinguish between Face-centred and end-centred unit cells.
What is the total number of atoms per unit cell in a face-centered cubic structure?
The number of atoms per unit cell in a body centered cubic structure is ____________.
Sodium metal crystallises in a body-centred cubic lattice with a unit cell edge of 4.29 Å. The radius of the sodium atom is approximately ______.
The empty space in the body-centered cubic lattice is ____________.
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:
The correct set of quantum numbers for 3d subshell is
The ratio of number of atoms present in a simple cubic, body-centred cubic and face-centred cubic structure are, respectively ______.
