Lithium metal has a body centred cubic structure. Its density is $0.53 g c m^{- 3}$ and its molar mass is $6.94 g m o l^{- 1}$ . Calculate the volume of a unit cell of lithium metal.

Open in App

Solution

Density of unit cell is given as
Density $- \frac{n o . o f a t o m p e r u n i t c e l l \times M o l a r m a s s o f a t o m}{M A \times V o l u m e o f u n i t c e l l}$
In BCC structure
at corner $= \frac{1}{8} \times 8 = 1$
at Body centre $+ 1 = 1 a t o m$
Total atom in BCC $= 2 a t o m$
Molar mass of lithium $= 6.94$
Given density $= 0.53 g c m^{- 3}$
On putting all values we have
$0.53 g c m^{- 3} = \frac{2 \times 6.94}{6.022 \times 10^{23} m o l^{- 1} \times 0.53 g c m^{- 3}} V = 4.34 \times 10^{23} c m^{3} V = 4.34 \times 10^{- 23} m l$
Thus volume of unit cell is $4.34 \times 10^{- 23} c m^{3}$

Suggest Corrections

Similar questions

0.53 g c m^{- 3}

6.94 g m o l^{- 1}

b c c

530 k g m^{- 3}

6.94 g m o l^{- 1}

N_{A} = 6.02 \times 10^{23} m o l^{- 1}

5.96 g / c m^{3}

(N_{0} = 6.023 \times 10^{23} a t o m s m o l^{- 1})

Join BYJU'S Learning Program