Question

lithium borohydride $\left( {{\text{LiB}}{{\text{H}}_4}} \right)$ crystallizes in an orthorhombic system with $4$molecules per unit cell. The unit cell dimensions are a=${\text{6}}{\text{.8}}{\buildrel _{\circ} \over {\mathrm{A}}}$, b=${\text{4}}{\text{.4}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$and c=${\text{7}}{\text{.2}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$. If the molar mass of the compound is $21.76\,{\text{g/mol}}$, the density (in ${\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$ ) of the compound will be:
A. $0.67$
B.$0.33$
C.$1.72$
D. none of theses

Answer 1

Hint: We should know the density formula to determine the answer. Density depends upon the number of atoms, mass and volume of a unit cell. Volume of the unit cell depends upon the dimension of the unit cell. The product of all dimensions gives the volume of an orthorhombic unit cell.

Formula used: ${\text{d}}\,{\text{ = }}\dfrac{{{\text{z}}\,{\text{m}}}}{{{{\text{N}}_{\text{a}}}{\text{V}}}}$

Complete step-by-step answer:
The formula to calculate the density of cubic lattice is as follows:
$d\, = \dfrac{{z\,m}}{{{N_a}{\text{V}}}}$
Where,
$d$ is the density.
$z$ is the number of atoms in a unit cell.
$m$ is the molar mass of the metal.
${N_a}$ is the Avogadro number.
${\text{V}}$ is the volume of a unit cell.
We know that the volume of an orthorhombic unit cell is the product of the dimensions of the unit cell. The unit cell dimensions are a=${\text{6}}{\text{.8}}{\buildrel _{\circ} \over {\mathrm{A}}}$, b=${\text{4}}{\text{.4}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$and c=${\text{7}}{\text{.2}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$. So, volume of orthorhombic unit cell is,
${\Rightarrow \text{1}}\,{\buildrel _{\circ} \over {\mathrm{A}}}\, = \,{10^{ - 10}}{\text{m}}$
$\Rightarrow {\text{V}}\,\, = {\text{6}}{\text{.8}} \times {\text{1}}{{\text{0}}^{ - 10}} \times {\text{4}}{\text{.4}} \times {\text{1}}{{\text{0}}^{ - 10}} \times {\text{7}}{\text{.2}} \times {\text{1}}{{\text{0}}^{ - 10}}$
$\Rightarrow {\text{V}}\,\, = {\text{215}}{\text{.424}} \times {\text{1}}{{\text{0}}^{ - 30}}\,{{\text{m}}^{\text{3}}}$
We will convert volume from meter to centimetre as follows:
$\Rightarrow 1\,{{\text{m}}^{\text{3}}}\, = \,{\left( {100} \right)^3}\,{\text{c}}{{\text{m}}^3}$
$\Rightarrow {\text{215}}{\text{.424}} \times {\text{1}}{{\text{0}}^{ - 30}}\,{{\text{m}}^{\text{3}}} = {\text{215}}{\text{.424}} \times {\text{1}}{{\text{0}}^{ - 24}}\,{\text{c}}{{\text{m}}^{\text{3}}}$
It is given that an orthorhombic system has $4$molecules per unit cell, so the number of atoms will be$4$.
Substitute $4$ for number of atoms, $21.76\,{\text{g/mol}}$ for molar mass of the compound, \[6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\]for Avogadro number, ${\text{215}}{\text{.424}} \times {\text{1}}{{\text{0}}^{ - 24}}\,{\text{c}}{{\text{m}}^{\text{3}}}$for unit cell volume.
$\Rightarrow {\text{d}}\, = \dfrac{{4 \times 21.76\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\, \times {\text{215}}{\text{.424}} \times {\text{1}}{{\text{0}}^{ - 24}}\,{\text{c}}{{\text{m}}^{\text{3}}}}}$
$\Rightarrow {\text{d}}\,{\text{ = }}\dfrac{{{\text{87}}{\text{.04}}\,{\text{g}}}}{{{\text{129}}{\text{.7}}\,{\text{c}}{{\text{m}}^{\text{3}}}}}$
$\Rightarrow {\text{d}}\,{\text{ = 0}}{\text{.67}}\,\,{\text{g/c}}{{\text{m}}^{\text{3}}}$
So, the density of the compound is$0.67\,\,{\text{g/c}}{{\text{m}}^3}$.

Therefore, option (A) $0.67$ is the correct answer.

Note:The value of the number of atoms depends upon the type of lattice. For face-centred cubic lattice, the number of atoms is four whereas two for body-centred and one for simple cubic lattice. In orthorhombic unit cell all dimensions are unequal ${\text{a}} \ne {\text{b}} \ne {\text{c}}$but angel in all faces are equal to ${\text{9}}{{\text{0}}^{\text{o}}}$ $\alpha = \beta = \gamma \,{\text{ = }}\,\,{\text{9}}{{\text{0}}^{\text{o}}}$. In FCC and BCC are dimensions are equal so, the volume of the unit cell is given as ${a^3}$ where, a is the unit cell length so the formula of density for FCC and BCC is, $d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$.