The body centered cubic cell of chronium has an edge lenth of 0.288nm. Calculte the density of chromium `(g//cm^(3))` : `("Atomic mass": Cr=52.0)`

Explanation

`M=52"g mol"^(-1)`
`d=7.20"g cm"^(-3)`,
`N_(A)=6.022xx10^(23)"mol"^(-1)`
z(b.c.c. lattice) = 2
`d=(ZxxM)/(a^(3)xxN_(A))`
`a^(3)=(ZxxM)/(d xxN_(A))`
`=(2xx52"g mol"^(-1))/(7.20"g cm"^(-3)xx6.022xx10^(23)"mol"^(-1))`
`=1040/(7.20xx6.022)xx10^(-24)"cm"^(3)`
`a^(3)=23.986xx10^(-24)"cm"^(3)`
`a=root(3)(23.986)xx10^(-8)"cm"`
`{:[("Let"" log"x=("log"23.986)^(1//3)=1/(3)"log"23.986),("log"x=1/(3)xx1.3800=0.4600),(x="antilog"0.46=2.89)]:}`
`thereforea=2.89xx10^(-8)"cm"`
`=2.89xx10^(-8)xx10^(10)"pm"`