Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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.
Distinguish between Face-centred and end-centred 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?
What is the coordination number of atoms in a cubic close-packed structure?
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Explain with reason sign conventions of ΔS in the following reaction
N2(g) + 3H2(g) → 2NH3(g)
Explain with reason sign conventions of ΔS in the following reaction
CO2(g) → CO2(g)
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?
Volume of unit cell occupied in face-centered cubic arrangement is ____________.
An element forms a cubic unit cell with edge length 405 pm. Molar mass of this element is 2.7 × 10−2 kg/mol and its density is given as 2.7 × 103 kg/m3. How many atoms of these elements are present per unit cell?
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?
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.
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 ______.
Edge length of unit cell of chromium metal is 287 pm with a bcc arrangement. The atomic radius is of the order:
The edge length of fcc cell is 508 pm. If the radius of cation is 110 pm, the radius of anion 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. |
Percentage of free space in body centred cubic unit cell 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 ______.
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.
The ratio of number of atoms present in a simple cubic, body-centred cubic and face-centred cubic structure are, respectively ______.
