Silver crystallises in a face centred cubic lattice with all the atoms at the lattice points. The length of the edge of the unit cell as determined by X-ray diffraction studies is found to be `4.077xx10^(-8)`cm. The density of silver is `10.5"g cm"^(-3)`. Calculate the atomic mass of silver.

Explanation

We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))orM=(rhoxxa^(3)xxN_(0)xx10^(-30))/(Z)`
According to available data :
Edge length (a) = `4.077xx10^(-8)cm=4.077xx10^(-8)xx10^(10)=407.7"pm"`
No. of atoms per unit cell (Z) = 4 , Density of silver = `10.50"g cm"^(-3)" "(because "fcc structure")`
Avogadro's Number (No.) = `6.022xx10^(23)"mol"^(-1)`
`therefore` Atomic mass of the element (M)
`=((10.50"g cm"^(-3))xx(407.7)^(3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)"cm"^(3)))/(4)=107.09"g mol"^(-1)`