Calculate the approximate number of unit cells present in 1 g of gold. Given that gold cyrstallises in a face centred cubic lathce (Given atomic mass of gold = 197 u).

Answer

`7.64xx10^(20)`

Explanation

Number of atoms in 1 g of gold `=("Avogadro's No. of atoms")/("Gram atom of gold")=((6.022xx10^(23)mol^(-1)))/(("197 g mol"^(-1)))xx(1g)`
Number of atoms per unit cell (fcc) of gold = 4
`therefore" Number of unit cells in 1 g of gold"=((6.022xx10^(23)mol^(-1))xx(1g)xx1)/(("197 g mol"^(-1))xx4)`
`=7.64xx10^(20)`