An element 'X' (At mass = 40 "g mol"^(-1)) having fcc structure, has unit cell length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g in 'X' (N_A=6.022xx10^23 "mol"^(-1))

An element 'X' (At mass = 40 "g mol"^(-1)) having fcc structure, has u

1.	An element 'X' (At mass = 40 "g mol"^(-1)) having fcc structure, has unit cell length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g in 'X' (N_A=6.022xx10^23 "mol"^(-1))
Answer» Solution :Given a=400 pm. For FCC, Z=4 `RHO=(ZxxM)/(a^3xxN_A)=(4xx40)/(6.022xx10^23xx(400xx10^(-10))^3)=4.15 g CM^(-3)` MASS of one unit cell =`40/(6.022xx10^23)xx4g` No. of unit cells in 4g of the element =`"Mass of the element "/"Mass of one unit cell"=4/((40xx4)//(6.022xx10^23))=1.505xx10^22`