An element forms a body centered cubic (bcc) lattice with edge length of 300 pm. If the density of the element is 7.2 g `cm^(-3)`, the number of atoms present in 324 g of it approximately is

Options

A. `3.33 times 10^(23)`
B. `6.66 times 10^(23)`
C. `3.33 times 10^(24)`
D. `6.66 times 10^(24)`

Answer

C

Explanation

Volume of bcc unit cell = 300 pm
`=(300 times 10^(-10)cm)^(3)=27 times 10^(-24)cm^(3)`
Volume of 324 g of the element `="Mass"/"Density"`
`" "=324/(7.2g" "cm^(-3))=45cm^(3)`
For a bcc structure, number of atoms per unit cell = 2
So, number of atoms `="Total volume"/"Volume of a unit cell"`
`" "=(2 times 45)/27 times 10^(24)=3.33 times 10^(24)`