Advertisements
Advertisements
प्रश्न
A substance crystallizes in fcc structure. The unit cell edge length is 367.8 pm. Calculate the molar mass of the substance if its density is 21.5 g/cm3.
उत्तर
Given: Edge length (a) = 367.8 pm = 3.678 × 10-8 cm, Density = 21.5 g/cm3
To find: Molar mass (M)
Formula: Density (ρ) = `"M n"/("a"^3 "N"_"A")`
Calculation: For an fcc lattice, number of atoms per unit cell is 4.
∴ n = 4
From formula,
Molar mass, M = `("a"^3 "N"_"A" rho)/"n"`
M = `((3.678 xx 10^-8)^3 "cm"^3 xx 6.022 xx 10^23 "atom mol"^-1 xx 21.5 "g cm"^-3)/(4 "atom")`
Calculation using log table:
`((3.678 xx 10^-8)^3 xx 6.022 xx 10^23 xx 21.5)/4`
`= ((3. 678)^3 xx 6.022 xx 2.15)/4`
= Antilog10 [3 × log10 3.678 + log10 6.022 + log10 2.15 – log10 4]
= Antilog10 [3 × 0.5657 + 0.7797 + 0.3324 – 0.6021]
= Antilog10 [1.6971 + 0.5100]
= Antilog10 [2.2071]
= 161.1
= 161.1 g mol−1
Molar mass of the substance is 161.1 g mol−1.
APPEARS IN
संबंधित प्रश्न
Answer the following in one or two sentences.
Which of the three types of packing used by metals makes the most efficient use of space and which makes the least efficient use?
Answer the following in one or two sentences.
Mention two properties that are common to both hcp and ccp lattices.
Answer the following in brief.
Calculate the packing efficiency of metal crystal that has simple cubic structure.
The density of iridium is 22.4 g/cm3. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.
In ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one-third of tetrahedral voids. What is the formula of the compound?
An element has a bcc structure with a unit cell edge length of 288 pm. How many unit cells and a number of atoms are present in 200 g of the element? (1.16 × 1024, 2.32 × 1024)
In case of hcp structure, how are spheres in first, second and third layers arranged?
A compound is formed by elements A and B. This crystallizes in the cubic structure when atoms A are at the comers of the cube and atoms B are at the centre of the body. The simplest formula of the compounds is ____________.
A compound of X and Y crystallizes in ccp structure in which the X atoms occupy the lattice points at the corners of cube and Y atoms occupy the centres of each of the cube faces. The formula of this compound is ____________.
The total volume occupied by particles m simple cubic unit cell is ____________.
The vacant space in simple cubic lattice is ____________.
Which of the following contains the highest number of atoms?
The percentage of vacant space of bcc unit cell is ____________.
Which among the following crystal structures the edge length of unit cell is equal to twice the radius of one atom?
1 mol of CO2 contains ____________.
Atoms of elements A and B crystallize in hep lattice to form a molecule. Element A occupies 2/3 of tetrahedral voids, the formula of molecule is ______.
Copper crystallizes as face centered cubic lattice, with edge length of unit cell 361 pm. Calculate the radius of copper atom.
Copper crystallises with fee unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper? (At. mass: Cu = 63.55 g mol-1)
AB is an ionic solid. If the ratio of ionic radius of A+ and B- is 0.52, what is the coordination number of B- ?
The relation between the radius of the sphere and the edge length in the body-centred cubic lattice is given by the formula ______.
Calculate packing efficiency in face-centred cubic lattice.
A compound made of elements C and D crystallizes in a fee structure. Atoms of C are present at the corners of the cube. Atoms of D are at the centres of the faces of the cube. What is the formula of the compound?
