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

`2.5xx10^(24)` atoms

Explanation

Step I. Calculation of mass of an atom of the elements
Let the mass of an atom of the element = m gram
Mass of the unit cell = `4xxm=(4m)g`
Edge length (a)= 400 pm = 400pm `xx10^(-10)` cm
Volume of the unit cell =`(400xx10^(-10)cm)^(3)=64xx10^(-24)cm^(3)`
Density of unit cell = 7 g `cm^(-3)`
Density of unit cell `=("Mass of unit cell")/("volume of unit cell")`
`("7 g cm"^(-3))=((4 m)g)/((64xx10^(-24)cm^(3)))`
`m=((7gcm^(-3))xx(64xx10^(-24)cm^(3)))/(4)=112xx10^(-24)g`
Step II. Calculation of the number of atoms in 280 g of the element.
`"No. of atoms"=("Mass of the element")/("Mass of one atom of the element")`
`=((280g))/((112xx10^(-24)g))=2.5xx10^(24)"atoms"`