An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contain `2 xx 10^(24)` atoms.

Explanation

Length of edge, `a = 250 pm = 250 xx 10^(12)` m
`= 250 xx 10^(-10) cm`
Volume of unit cell = `(250 xx 10^(-10) cm)^(3)`
`= 15.625 xx 10^(-24) cm^(3)`
Mass of unit cell = No. of atoms in unit cell `xx` Mass of each atom
Since the element has fcc arrangement, the number of atoms per unit cell, Z = 4
Mass of an atom = `300/(2 xx 10^(24)) g`
`:.` Mass of unit cell - `(300)/(2 xx 10^(24)) xx 4 = 6.0 xx 10^(-22) g`
Density of unit cell = `("Mass of unit cell")/("Volume of unit cell")`
`= (6.0 xx 10^(-22) g)/(15.625 xx 10^(-24) cm^(3))`
`= 38.4 g cm^(-3)`