Question

Sodium has a bcc structure with nerast neighbour distance $365.9pm$ Calculate its density. (Atomic mass of $Na=23$)

• A. $2.44\times {10}^{31}g/cc$
• B. $3.64\times {10}^{32}g/cc$
• C. $3.44\times {10}^{33}g/cc$
• D. $1.51\times {10}^{31}g/cc$

Answer 1

The correct option is A $2.44\times {10}^{31}g/cc$

$d=\frac{Z\times Mwt}{N_a\times a^3}$

$a=\frac{4r}{\sqrt{3}}$ for $bcc$

$a=\frac{4\times 365.9\times {10}^{-12}}{\sqrt{3}}=1014.8\times {10}^{-12}$

So $d=\frac{Z\times Mwt}{N_a\times a^3}$ or $d=\frac{2\times 23}{6\times {10}^{23}\times (1014.8\times {10}^{-12}}{)}^3=2.44\times {10}^{31}$