An element (at mass 60) havind FCC structure has a density of `6.23"g cm"^(-3)`. What is the edge length of the unit cell ?

An element 'X' with an atomic mass of 60g/ mol has density of 6.23"g cm"^(-1) . If the edge length of its unit cell is 400 pm, identity the type of cell. Also calculate the radius of an atom of this element.

Density of a unit cell is respresented as rho = (" Effective no. of atoms (s)" xx "Mass of a unit cell ")/("Volume of a unit cell ")=(Z.M)/(N_(A).a^(3)) where , mass of unit cell =mass of effectuive no . of atoms(s) or ion (s). M= At . mass// formula N_(A) = Avogadro' s no . rArr 6.0323 xx 10^(23) a= edge lemght of unit cell Silver crystallizes in a fcc lattice and has a density of 10.6 g//cm^(3) . What is the length of a edge of the unit cell ?

The compound CuCl has fcc structure like ZnS. Its density is 3.4 gcm^-3 . What is the length of the edge of the unit cell?

The CsCl has simple cubic sturcute and Its density is 3.99 g cm^-3 . Calculate the edge length of unit cell.

Silver having atomic mass 107.87 crystallizes in faces centred cubic structure. If density is 10.76"gcm"^(-3) calculate the edge length of unit cell.

An element A crystallines in fcc structure. 200 g of this element has 4.12xx10^24 atoms. The density of A is 7.2 gcm^-3 . Calculate the edge length of the unit cell.

Caesium chlorde cystallised as cubic lattices and has a density of 4.0 gcm^-3 . Calculate the length of the edge of the unit cell of caesium chloride.( Molar mass of CsCl=168.5 gmol^-1 ).

The unit of an element of atomic mass 96 and density 10.3gcm^-3 is a cube with edge length of 314 pm. Find the structure of the crystal lattice (simple cubic, FCC or BCC) (Avagadro's constant, N_0=6.023xx10^23 mol^-1 ).

A metal having atomic mass 50 g "mol"^(-1) has a body centred cubic crystal structure. The density of the metal is 596 g "cm"^(-1) . Find the volume of unit cell ?