A substance has bcc structure with edge length of 288 pm . the density of the element is 7.2 g cm2 . how many atoms r present in 2208g of the element ?
Ï = ZM / a³N
Z = 2 for Bcc
Ï = 7.2
N the Avrogadro constant = 6.022x10²³
a the cubic unit cell lattice parameter = 288 ppm = 288 x 10-12 m
using formula
7.2 = 2 M / ( 288 x 10-12) 3 6.022x10²³
M = 51.9 gm
Element is Chromium
Total weight of element 2208
No of moles = 2208/ 51.9 = 42.5 moles = 42.5 x 6.022 x 1023 = 2.56 x 1025 atoms