Question

tungsten has a density of 19.35 g cm${}^{-3}$ in its bcc structure and the length of the unit cell side is 316 pm. How many atoms does 50 gm. of element contain?

Answer 1

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$Z$ for BCC $=8+\frac{1}{8}+1=2$

$M_A=?$

$a=316$ $pm$

$19.35=\frac{2\times M_A}{6.023\times {10}^{23}\times (316{)}^3\times {10}^{-30}}$

$M_A=183.84$

Moles of Tungsten $=\frac{50}{183.84}=0.272$ moles

No. of atoms $=0.272\times 6.023\times {10}^{23}=1.64\times {10}^{23}$ atoms