Question

Lithium borohydride, LiBH4, crystallizes in an orthorhombic system with 4 molecules per unit cell.

The unit cell dimensions are : a = 6.81 Armstrong , b = 4.43 Armstrong and c = 7.17 Armstrong. Calculate the density of the
crystal. Take atomic mass of Li = 7, B = 11 and H = 1 a.m.u.
​

Answer 1

Answer :

• Density of the unit cell = $\bold{0.676\:g\:cm^{-3}}$

Step-by-step explanation :

Given that,

• Dimensions of unit cell, a = 6.81 Å = $\bold{6.81\times{}10^{-8}\:cm}$
• b = 4.43 Å = $\bold{4.43\times{}10^{-8}\:cm}$
• c = 7.17 Å = $\bold{7.17\times{}10^{-8}\:cm}$

• Also, atomic mass of, Li = 7 a.m.u.
• B = 11 a.m.u.
• H = 1 a.m.u.

$\hrulefill$

Now, molar mass of,

$\bold{LiBH_4=7+11+1\times{}4=22\:g\:mol^{-1}}$

∴ Mass of the unit cell = $\bold{\frac{4\times{}22\:g\:mol^{-1}}{6.022\times{}10^{23}\:mol^{-1}}}$

$\implies\bold{14.62\times{}10^{-23}\:g}$

Now, Volume of the unit cell = a × b × c

$\implies\bold{(6.81\times{}10^{-8}\:cm)(4.43\times{}10^{-8}\:cm)(7.17\times{}10^{-8}\:cm)}$

$\implies\bold{21.63\times{}10^{-23}\:cm^3}$

$\hrulefill$

∴ Density of the unit cell = $\bold{\dfrac{Mass}{Volume}}$

$\implies\bold{\dfrac{14.62\times{}10^{-23}\:g}{21.63\times{}10^{-23}\:cm^3}}$

$\implies\bold{0.676\:g\:cm^{-3}}$