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An element 'x' (Atomic mass = 40 g mol^(...

An element 'x' (Atomic mass = 40 g `mol^(-1)`) having f.c.c. structure has unit cell edge length of 400 pm. Calculate the density of 'x' and the number of unit cells in 4 g of 'x'

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Calculation of density of unit cell.
Density of unit cell `=(ZxxM)/(N_(0)xxa^(3))`
According to available data :
Edge length (a) = 400 pm = 400
Atomic mass X (M) = 40 g `mol^(-1)`
No. of atoms in the unit cell (Z) = 4
Avogadro's number `(N_(0))=6.022xx10^(23)"mol"^(-1)`
`"Density of unit cell"=(4xx("40 g mol"^(-1)))/((6.022xx10^(23)"mol"^(-1))xx(400)^(3)xx(10^(-30)"cm"^(3)))="4.15 g cm"^(-3)`
Step-II. No. of unif cells in 4 g of X.
Mass of the unit cell `=(ZxxM)/(N_(0))=(4xx("40 g mol"^(-1)))/((6.022xx10^(23)"mol"^(-1)))`
`"No. of unit cells in 4 g of 'X'"((6.022xx10^(23)"mol"^(-1)))/((4xx"40 g mol"^(-1)))xx(4 g)=1.505xx10^(22)`
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