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Sodium has a `bc c` structure with nearest neighbour distance of `365.9` pm. Calculate its density. (Atomic mass of sodium `= 23`)A. `1.5g//cm^(3)`B. `3g//cm^(3)`C. `4g//cm^(3)`D. `5g//cm^(3)`

Answer» For the bcc structure, nearest neighbour distance (d) is related to the edge (a) as `d=sqrt(3)/(2)a `
or `a=(2)/sqrt(3)d=(2)/(1.732)xx365.9=422.5`pm
For bcc structure, `Z_(eff)=2`
For sodium .Aw=23
`therefore r=(Z_(eff)xxA_(w))/(a^(3)xxN_(A)xx10^(-30))=(2xx23)/((422.5)^(3)xx(6.02xx10^(23))xx10^(-23))gcm^(-3)`
`=1.51g cm^(-3)`


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