Question

Tungsten has a density of $19.35$ g $c{m}^{-3}$ and length of the side of the unit cell is $316$ pm. The unit cell has a body-centred unit cell. How many atoms of the element does $50$ g of the element contain?

Answer 1

Length of edge of the unit cell $=316pm=316\times {10}^{-10}cm$

Volume of the unit cell $=(316\times {10}^{-10}{)}^3$

$=3.2\times {10}^{-23}c{m}^3/unitcell$

$Density=\frac{Atomicmass\times Z}{Unitcellvolume\times N_A}$

$Atomicmass=\frac{Density\times Unitvolume\times N_A}{Z}$

$=\frac{19.35\times 3.2\times {10}^{-23}\times 6.023\times {10}^{23}}{2}$

$=186.5g/mol$

As 186.5 g the element contains $N_A$ atoms $=6.023\times {10}^{23}atoms$

So, 50 g the element contains $N_A$ atoms $=6.023\times {10}^{23}\times \frac{50}{186.5}$

$=1.614\times {10}^{23}atoms$