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`).

Calculate the density of NaCl if edge length of Nacl unit cell is 564 pm. (Molar mass of NaCl=58.5 g mol^-1)

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

Chromium crystallizes in a body centred cubic lattice whose density is 7.20g/cm^3 . The length of the edge of the unit cell is 288.4 pm. Calculate the Avogadro number (atomic mass of chromium=52).

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 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.

A metal crystallizes in a face centred cubic unit cell with a=0.560nm. Calculate the density of the metal if contains 0.1% Schottky defects. (Atomic mass of metal = 40 g mol^-1 ).

A face centered cubic element (atomic mass =60) has edge length of 400pm. What is the density of unit cell

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 ?

Silver crystallines in fcc lattice. If edge length of the cell is 4.077xx10^-8cm and density is 10.5 g cm^-3 . Calculate the atomic mass of silver.