Advertisements
Advertisements
प्रश्न
An element with molar mass 27 g/mol forms a cubic unit cell with edge length of 405 p.m. If the density of the element is 2.7 g/cm3, what is the nature of the cubic unit cell?
Advertisements
उत्तर
Given:
Edge length (a) = 405 pm = 4.05 × 10−8 cm
Molar mass = 27 g mol−1
Density (ρ) = 2.7 g/cm3 = 2.7 g cm−3
To find: Nature of cubic unit cell
Formula: Density (ρ) = `(M n)/(a^3 N_A)`
Calculation: From the formula,
Density, ρ = `(M n)/(a^3 N_A)`
∴ `2.7 "g cm"^-3 = (27 "g mol"^-1 xx "n")/((4.05 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1)`
∴ n = `(2.7 "g cm"^-3 xx (4.05 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1)/(27 "g mol"^-1)`
= 4.00
∴ Number of atoms in unit cell = 4
Since the unit cell contains 4 atoms, it has a face-centred cubic (fcc) or ccp structure.
The nature of the given cubic unit cell is face-centred cubic (fcc) or ccp unit cell.
संबंधित प्रश्न
Obtain the relationship between the density of a substance and the edge length of the unit cell.
Write the relationship between radius of atom and edge length of fcc unit cell.
Give the percentage of empty space in bcc lattice.
Calculate the number of unit cells in 0.3 g of a species having density of 8.5 g/cm3 and unit cell edge length 3.25 × 10-8 cm.
Derive the relationship between density of substance, its molar mass, and the unit cell edge length. Explain how you will calculate the number of particles, and a number of unit cells in x g of metal.
The pyranose structure of glucose is ____________.
An element (atomic mass M g/mol) having bcc structure has unit cell edge 400 pm. The density of the element is ____________ g/cm3.
[NA = 6.0 × 1023 atom mol−1)
How many total constituent particles are present in simple cubic unit cell?
The percentage of unoccupied volume in simple cubic cell is ______.
The number of atoms in 500 g of a fcc crystal of a metal with density d = 10 g/cm3 and cell edge 100 pm, is equal to ____________.
If the edge of a body-centred unit cell is 360 pm, what will be the approximate radius of the atom present in it? (in pm)
An element crystallizes bcc type of unit cell, the density and edge length of unit cell is 4 g cm−3 and 500 pm respectively. What is the atomic mass of an element?
An element with density 2.8 g cm−3 forms fcc unit cell having edge length 4 × 10−8 cm. Calculate molar mass of the element.
Copper and silver have ____________ crystal structure.
Consider the following unit cell.
The number of particles (spheres) per unit cell is:
Which of the following formulae is used to find edge length of bee unit cell?
A metal has an fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm−3. The molar mass of the metal is ______.
(NA Avogadro's constant = 6.02 × 1023 mol−1)
At room temperature, polonium Crystallises in a primitive cubic unit cell. If a = 3.36 Å. Calculate the theoretical density of polonium. [It's atomic weight is 209 g/mol.]
Silver crystallizes in the fcc structure. If the edge length of the unit cell is 400 pm, calculate the density of silver (Atomic mass of Ag = 108).
An element crystallises in fee structure. If molar mass of element is 72.7 g mol -1, the mass of its one unit cell will be ______.
The number of particles present in Face Centred Cubic Unit cell is/are ______.
The cubic unit cell of a metal (molar mass = 63.55g mol−1) has an edge length of 362 pm. Its density is 8.92g cm−3.
The type of unit cell is ______.
Calculate the molar mass of an element having a density of 19.2 g cm−3 if it forms an fcc structure [a3 × NA = 40 cm3 mol−1].
