an element adopts a cubical crystal structure in which only 68% ocupied.Edge length is 288 pm.If density of element is 7.2 g/cm3 ,the number of atoms in 144g of element ???
Volume=2.4*10^(-23) cc
106This is BCC.. (68%)
so no. of atoms per unit cell = 2
density = 2*M/V where M = mass of one atom.
knowing mass of one atom,
you can calculate no. of atoms in 144g,
btw why have they given edge length as well as volume.. is it volume of some other thing?
19info about packing fraction :
simple or primitive unit cell -- 52.36%
body centred cubic cell--68.02%
face centred cubic cell--74.06%
24"btw why have they given edge length as well as volume.. is it volume of some other thing?"
the same is my dbt....so posted the ques here.. ....why excess info ??
19V=abc(1- cos2α -cos2β -cos2γ +2cos2α cos2β cos2 γ)1/2
for a cubic cell, a=b=c and α=β=γ
...=> V=abc