An element crystallizes as body `-` centred cubic lattic. Its density is `7.12g cm^(-3` and the length of the side of the unit cell is `2.88Å`. Calculate the number of atoms present is `288g` of the element.

Options

A. `1.693 X 10^(24)`
B. `1.693 X 10^(23)`
C. `3.386 X 10^(23)`
D. `3.386 X 10^(24)`

Answer

D

Explanation

Volume of unit cell
`=a^(3)=(2.88overset(@)A)=(2.88xx10^(-6)cm)^(3)`
`=23.887xx10^(-24)cm^(3)`
Volume of 288g of the element
`=("Mass")/("Density")=(288g)/(7.12g cm^(-3))=40.449cm^(3)`
`=("volume of the element")/("volume of the unit cell")=(40.449)/(23.887xx10^(-24))=1.693xx10^(24)`
`because` Each unit cell of b.c.c. lattice contains 2 atoms
`therefore` number of atoms in 288g of element
=No. of unit cells xx no. of atoms per unit cell
`=1.693xx10^(24)xx2=3.386xx10^(24)`