An element has a face centered cubic unit cell with a length of 352.4pm along and edge . The density of the element is `8.9 g cm^(-3)` . How many atoms are present in 100 g of an elements ?

Explanation

The formula to be used is
` M = (rho xx a^(3) xx N_(A))/(2)`
`rho ` = density of the crystal ` = 8.9 g cm ^(-3)`
a = edge length = 352.4 pm
=352.4 xx ` 10 ^(-10) cm`
`N_(A) ` = Avogadro number =`6.023 xx 10 ^(23)`
Z = Number of atoms is a fcc .
unit cell. 4
Substituting .
M = Molar mass of the atom.
` M = (8.9 xx (352.4 xx 10^(-10) )^(3) xx 6.023 xx 10^(23))/(4)`
`= 58.65 g mol^(-1)`
58.65g of the substance contains
`=6.023 xx 10^(23)` atoms
100 g of the substance will contain
` = (6.023 xx 10^(23)) /(58.65) g mol^(-1)`
58.65 g of the substance contains
` = 6.023 xx 10^(23) ` atoms
100g of the substance will contains
= `(6.023 xx 10^(23))/(58.65)xx 100`
` = 10.23 xx 10^(23)` atoms .