Advertisements
Advertisements
प्रश्न
A unit cell of iron crystal has edge length 288 pm and density 7.86 g.cm-3. Find the number of atoms per unit cell and type of the crystal lattice.
Given : Molar mass of iron = 56 g.mol-1; Avogadro's number NA = 6.022 x 1023.
Advertisements
उत्तर
Edge length of unit cell (a) = 288pm
volume of unit cell = (a)3
=(288)3 = 2.389 × 10-23 cm3
density of iron = ρ = 7.86g.cm3
Mass of iron unit cell (M) = Density × Volume
= 7.86 × 2.389 × 10-23
= 18.778 × 10-23 g
molar mass of iron = 56 g.mol-1
moles of iron in a unit cell = mass of iron in unit cell / molar mass of iron
= 18.778 × 10-23 / 56
= 3.353 × 10-24 mol
number of atoms per unit cell = moles of iron in a unit cell × Avogadro's number
= 3.353 × 10-24 × 6.022 x 1023
= 2.0191
≈ 2 atoms per unit cell
In a body centered cubic structure (bcc), the total number of atoms equals 2.
APPEARS IN
संबंधित प्रश्न
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)
Distinguish between Hexagonal and monoclinic unit cells
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?
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
A face centred cube (FCC) consists of how many atoms? Explain
An element has atomic mass 93 g mol−1 and density 11.5 g cm–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell. (NA = 6.023 × 1023 mol−1)
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)
The density of silver having an atomic mass of 107.8 g mol- 1 is 10.8 g cm-3. If the edge length of cubic unit cell is 4.05 × 10- 8
cm, find the number of silver atoms in the unit cell.
( NA = 6.022 × 1023, 1 Å = 10-8 cm)
Number of types of orthorhombic unit cell is ___________.
What is the total number of atoms per unit cell in a face-centered cubic structure?
TiCl has the structure of CsCl. The coordination number of the ions in TiCl is ____________.
A substance forms face-centered cubic crystals. Its density is 1.984 g/cm3 and the length of the edge of the unit cell is 630 pm. Calculate the molar mass in g/mol?
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).
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 ______.
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 ______.
Match the type of unit cell given in Column I with the features given in Column II.
| Column I | Column II |
| (i) Primitive cubic unit cell | (a) Each of the three perpendicular edges compulsorily have the different edge length i.e; a ≠ b ≠ c. |
| (ii) Body centred cubic unit cell | (b) Number of atoms per unit cell is one. |
| (iii) Face centred cubic unit cell | (c) Each of the three perpendicular edges compulsorily have the same edge length i.e; a = b = c. |
| (iv) End centred orthorhombic cell | (d) In addition to the contribution from unit cell the corner atoms the number of atoms present in a unit cell is one. |
| (e) In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three. |
The correct set of quantum numbers for 3d subshell is
The correct set of quantum numbers for 2p sub shell 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.
