Each edge of a cubic unit cell is 400pm long. If atomic mass of the elements is 120 and its desity is `6.25g//cm^(2)`, the crystal lattice is: `(use N_(A)=6 xx 10^(23))`

In a cubic lattice each edge of the unit cell is 400 pm. Atomic weigth of the element is 60 and its density is 625g//c.c. Avogadro number = 6times10^(23) .The crystal lattice is

In a cubic lattice each edge of the unit cell is 400 pm. Atomic weigth of the element is 60 and its density is 625g//c.c. Avogadro number = 6times10^(23) .The crystal lattice is

The density of an cubic unit cell is 10.5 g cm^(-3) . If the edge length of the unit cell is 409 pm. Find the structure of the crystal lattice (Atomic mass = 108).

An element has a body centered cubic structure with a unit cell edge length of 400 pm. Atomic mass of an element is 24 gmol^(-1) What is the density of the element? (N_A = 6 times 10^23 mol^-1)