Question

An element has a body-centred
cubic (bcc) structures with a cell
edge of 444 pm. The density of the
element is 7.2 g/cm3. How many
atoms are present in 404 g of the
element?​

Answer 1

Explanation:

Volume of the unit cell=(444×10^-10)³

8.75×10^-23Cm³

density= mass / volume

Density of cubic unit cell -

Z×mass of 1 atom

volume of unit cell

7.2= 2×404

8.75×10^-23 × N

N= 808

7.2×8m75×10^-23

N= 12.82×10^23

- wherein

Where,

d = density of crystal

z = no. of effective constituent particles in one unit cell

M = molecular weight

a = edge length of unit cell

No = 6.022*1023

hence the answer is 12.82×10^23..........

i hope it was helpful to u guys .......

thank you......