An element with density `11.2 g cm^(-3)` forms a f. c. c. lattice with edge length of `4xx10^(-8)` cm. Calculate the atomic mass of the element. (Given : `N_(A)= 6.022xx10^(23) mol^(-1)`

Answer

`107.9 u`

Explanation

`rho = (Z xx M)/(a^3 xx N_A)`
`Z = 4 ("fcc"), a = 4 xx 10^(-8) cm, rho = 11.2 g cm^(-3)`
`11.2 = (4 xx M)/((4 xx 10^(-8))^(3) xx (6.02 xx 10^(23))`
or `M = ((11.2) xx (4 xx 10^(-8))^(3) xx (6.02xx 10^(23)))/(4)`
`= 107.9 u`.