Advertisements
Advertisements
प्रश्न
Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, calculate the atomic mass of silver
उत्तर
It is given that the edge length, a = 4.077 × 10−8 cm
Density, d = 10.5 g cm−3
As the lattice is fcc type, the number of atoms per unit cell, z = 4
We also know that, NA = 6.022 × 1023 mol−1
Using the relation:
`d = (zM)/(a^3N_A)`
`=> M = (da^3N_A)/z`
`= (10.5 "gmc"^(-3)xx(4.077xx10^(-8) cm)^3xx6.022xx10^23 mol^(-1))/4`
= 107.13 gmol−1
Therefore, atomic mass of silver = 107.13 u
APPEARS IN
संबंधित प्रश्न
Silver crystallises in FCC structure. If density of silver is 10.51 gcm-3, calculate the volume of unit cell. [Atomic mass of slive (Ag) = 108 gm-1]
Niobium crystallises as body centred cube (BCC) and has density of 8.55 Kg / dm-3 . Calculate the attomic radius of niobium.
(Given : Atomic mass of niobium = 93).
An element with density 10 g cm−3 forms a cubic unit cell with edge length of 3 × 10−8 cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol−1?
An element crystallizes in a f.c.c. lattice with a cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 × 1024 atoms.
An element with density 2.8 g cm–3 forms a f.c.c. unit cell with edge length 4 x 10–8 cm. Calculate the molar mass of the element.
(Given : NA = 6.022 x 1023 mol –1)
An element with density 11.2 g cm–3 forms a f.c.c. lattice with edge length of 4 × 10–8 cm.
Calculate the atomic mass of the element.
(Given : NA = 6.022 × 1023 mol–1)
Represent a cell consisting of Mg2+ | Mg half cell and Ag+
| Ag half cell and write the cell reaction.
( `"E"_("Ag")^° = 0.799 "V", "E"_("Mg")^° = - 2.37 "V"`)
An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. calculate the number of atoms in 108 g of the element.
A unit cell of NaCl has 4 formula units. Its edge length is 0.50 nm. Calculate the density if molar mass of NaCl = 58.5 g/mol.
Tetragonal crystal system has the following unit cell dimensions
Crystalline CsCl has density 3.988 g cm-3. The volume occupied by single CsCl pair in the crystal will be ______.
(Molar mass CsCl = 168.4 g/mol)
A unit cell of sodium chloride has four formula units. The edge length of the unit cell is 0.564 nm. The density of sodium chloride is ______ g/cm3.
