An element crystallizes in f.c.c. lattice with edge length of 400 pm. The density of the element is 7 g `cm^(-3)`. How many atoms are present in 280 g of the element ?

Answer

`2.5 xx 10^(24)` atoms

Explanation

Edge length = 400 pm =`400 xx 10^(-10) cm`
Volume of unit cell = `(400 xx 10^(-10))^(3) = 64 xx 10^(-24) cm^3`
Volume of 280g of element = `(280 g)/(7 g cm^(-3)) = 40 cm^(3)`
Number of unit cells in this volume
`= (40)/(64 xx 10^(-24)) = 6.25 xx 10^(23)`
For fcc, number of atoms per unit cell = 4
`:.` No of atoms in 280 g of element = `6.25 xx 10^(23) xx 4`
`= 2.5 xx 10^(24)` atoms .