Question

In an orthorhombic unit cell, the value of a, b and c are respectively \[{\rm{4}}{\rm{.2}}\,\mathop {\rm{A}}\limits^{\rm{o}} \], \[8.6\,\mathop {\rm{A}}\limits^{\rm{o}} \] and \[8.3\,\mathop {\rm{A}}\limits^{\rm{o}} \]. Given the molecular mass of the solute is \[{\rm{155}}\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]and that of density is \[\rm{3.3 \,gm}/\rm{cc}\], the number of formula units per unit cell is
(A) \[2\]
(B) \[3\]
(C) \[4\]
(D) \[{\rm{6}}\]

Answer 1

Hint: As we know that, the density is measured by the packing of the atoms in the unit cell, the total number of formula unit cells tells us how many formulae can be generated in the given unit cell.

Complete step by step answer
Formula unit is generated by the total number of ions (Cation or anions) in the unit cell such as in rock salt, the number of cation per unit cell is \[4\] and the number of anions per unit cell is \[4\] so, therefore, the total number of formula per unit cell will be \[4\].
Density of the orthorhombic unit cell is calculated by the formula-
\[{\rm{d = }}\dfrac{{{\rm{Z}}\,{\rm{ \times M}}\,\,}}{{{{\rm{N}}_{\rm{a}}}{\rm{ \times }}\,{\rm{a \times b}}\,{\rm{ \times }}\,{\rm{c}}}}\]
Where, \[{\rm{Z}}\] is the number of ions per unit cell, \[{\rm{M}}\]is the molecular mass of the compound, \[{{\rm{N}}_{\rm{a}}}\]is the Avogadro constant and a, b and c are the edge length.
In this question we are given as
\[{\rm{M}}\,{\rm{ = }}\,{\rm{155}}\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]
\[{\rm{a}} = {\rm{4}}{\rm{.2}}\,\mathop {\rm{A}}\limits^{\rm{o}} = {\rm{4}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\,{\rm{cm}}\]
\[{\rm{b}} = {\rm{8}}{\rm{.6}}\,\mathop {\rm{A}}\limits^{\rm{o}} = {\rm{8}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\,{\rm{cm}}\]
\[{\rm{c}} = {\rm{8}}{\rm{.3}}\,\mathop {\rm{A}}\limits^{\rm{o}} = {\rm{8}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\,{\rm{cm}}\]
$\rm{d=3.3\, gm}/\rm{cc}$
Now we are left with value of
\[{\rm{Z}}\,{\rm{ = }}\,{\rm{?}}\]
By putting above values in the above equation, we find as
${\rm{Z}}$ ${\rm{ = }}\,$ $\dfrac{6.022\times{10}^{23} {\rm{mol}^{-1}} \times {\rm{4}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\,{\rm{cm}}\times{\rm{8}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\,{\rm{cm}} \times {\rm{8}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\,{\rm{cm}} \rm{d=3.3\, gm}/\rm{cc}}{155{\rm{gmol}}^{-1}} $
${\rm{Z}}$ ${\rm{ = }} \rm{3.84} $
${\rm{Z}} {\rm{ \simeq }} \rm{4}$

Therefore, the correct option is C. 4

Note:
While calculating the density or numerical question, we generally forget unit conversions, that make us very disappointed, so, try to solve numerical with units.
In the thermodynamics and solution chapter, there are more unit conversions, so try to practice more numerically.