Sodium has a b.c.c. structure with nearest neighbour distance 365.9 pm, calculate its density. (Atomic mass of sodium = 23)

Explanation

For a b.c.c. lattice.
`sqrt(3)a=2d`
`a=2/(sqrt(3))d=2/(sqrt(3))xx365.9`
`=731.8/(3)xxsqrt(3)=243.9xx1.732=422.5 pm`
Density `d=(Mxxz)/(a^(3)xxN_(A))`
Here `M=23"g mol"^(-1),z=2` (for bcc unit cell)
a = 422.5 pm
`=422.5xx10^(-10)cm`,
`N_(A)=6.022xx10^(23)"mol"^(-1)`
`therefored=(23"g mol"^(-1)xx2)/((422.5xx10^(-10)"cm")^(3)xx(6.022xx10^(23)"mol"^(-1))`
`=(46xx10)/((4.225)^(3)xx6.022)"g/cm"^(3)`
`=460/(75.42xx6.022)"g/cm"^(3)`
`=460/(454.17)"g/cm"^(3)`
`=1.013"g/cm"^(3)`