Question

An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unit cell is `24xx10^(-24) cm^3` and density of element is `7.2 g cm^(-3)` , calculate the number of atoms present in 200 g of the element.

Answer 1

No. of atoms per unit cell =`8xx1/8+2=3` i.e., Z=3
Volume of the unit cell =`a^3 = 24xx10^(-24) cm^3`
`rho=(ZxxM)/(a^3xxN_0)`
`therefore 7.2=(3xxM)/((24xx10^(-24))xx(6.023xx10^23))` or M=34.69
Thus , 34.69 g of the element have atoms =`6.023xx10^23`
`therefore` 200 g of element will have atoms =`(6.023xx10^23)/34.69xx200=3.4722xx10^24` atoms