Advertisements
Advertisements
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)
Advertisements
उत्तर
Unit cell of FCC `= 1/8 xx 8 + 6 xx 1/2`
= 4 atoms
Mass of unit cell of FCC `= 4 xx 197/(6.022 xx 10^23)`
= 130.85 × 10−23 g
`rho = 19.3` g/cm−3
Volume of unit cell = `(130.85xx10^-23)/19.3`
= 6.78 × 10−23 cm3
a3 = 6.78 × 10−23
a is edge of unit cell
`a = root(3)(6.78 xx 10^(-23)`
= 4.08 × 10−8 cm
For FCC
`a = sqrt8.r`
`r = a/sqrt8`
= `(4.08 xx 10^(-8))/sqrt8`
= 1.44 × 10−8 cm
= 144 pm
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?
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?
What is the coordination number of atoms in a cubic close-packed structure?
Explain with reason sign conventions of ΔS in the following reaction
N2(g) + 3H2(g) → 2NH3(g)
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)
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?
The number of atoms per unit cell in a body centered cubic structure is ____________.
A metal has a body-centered cubic crystal structure. The density of the metal is 5.96 g/cm3. Find the volume of the unit cell if the atomic mass of metal is 50.
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).
Gold has a face-centered cubic lattice with an edge length of the unit cube of 407 pm. Assuming the closest packing, the diameter of the gold atom 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 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:
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 coordination number for body center cubic (BCC) system is
If a represents the edge length of the cubic systems, i.e. simple cubic, body centred cubic and face centered cubic, then the ratio of the radii of the sphere in these system will be:-
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 ______.
