Question

An element having FCC lattice structure has a cell edge of 120 pm . The density of the element is 6.8g/cm^3
How many atoms are present in 408 g of the element?​

Answer 1

Answer:

density=mass/volume×z/Na

Explanation:

volume (a^3), z(4),Na(6×10^23),m(408) put the values solve it and get the answer