Question

An element crystallizes in fec lattice with a cell edge of 300 pm. The density of the element is 10.8 g `cm^(-3)`. Calculate the number of atoms in 108 g of the element.

Answer 1

For f.c.c structure z =4
`a=300 "pm"=300x10^(-10)cm,d=10.8cm^(-3)`
`d=(zxxM)/(a^(3)xxNA)`
`M=(dxxa^(3)xxNA)/z=(10.8xx(300xx10^(-10)^(3)xx6.022xx10^(23)))/(4)=43.9 gmol^(-1)`
`because` 43.9 g of element contains = `6.022xx10^(23)` atoms
108 g of the elements contains `=(6.022xx10^(23))/43.9xx108=14.81xx10^(23)` atoms