An element has a face centered cubic unit cell with a length of 353.4 pm along an edge. The density of the element is 8.9 `g cm^(-3)`. How many atoms are present in 100g of an element?

Explanation

Given `:`
Density `= 8.9 g cm^(-3)`
Number of Atom in unit cell of fcc =4
Edge length `= 352.4 p m =352.4 xx 10^(-10) cm`
Avogadro's number `= 6.022 xx 10^(23) mol^(-1)`
Mass of element 100g
`p =(Z xx M )/(N_(A) xxa^(3))`
`8.9 g cm^(-3) = ( 4 xx M )/( (6.022 xx 10^(23) mol^(-1) ) xx( 352.4 xx 10^(-10)cm))`
`M = 58.6 g //`mole
Moles of element `= ( "Mass of element")/("Molar mass of element")`
`= (100g)/( 58.6 g//mol)`
=1.7 moles.
As, 1 mole contains `6.022 xx 10^(23) ` number of atoms.
So, 1.7 mole contains `= 1.7 xx ( 6.022 xx 10^(23))`
`= 10.23 xx 10^(23)` number of atoms.
`:.` The number of atoms present in 100g of element is `10.23 xx 10^(23)`.