Question

Calculate the number of unit cells in ${\rm{8}}{\rm{.1 g}}$ of aluminum if it crystallizes in a face-centred cubic (f.c.c.) structure? (Molar mass of ${\rm{Al}} = {\rm{27 g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$)
Given:
a. Mass of aluminum is: $w = {\rm{8}}{\rm{.1 g}}$
b. Type of the structure is: face-centred cubic
c. Atomic mass of aluminum is:$M = {\rm{27 g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$)

Answer 1

The number of atoms in given mass can be determined by using the molar mass and Avogadro number that in turn can be related to the number of atoms per unit cell.

Complete step by step solution:
We know that the molar mass of a substance is the mass of ${\rm{1 \,mole}}$ of that substance. Mathematically, we can express it as per the following unit formula:$\dfrac{{{\rm{1 \,mole}}}}{{{\rm{molar \,mass\, in\, g}}}}$
For aluminum, we can write the unit formula by using the given molar mass as follows:
$\dfrac{{{\rm{1 \,mole\, Al}}}}{{{\rm{27\, g\, Al}}}}$
We can find out the number of moles of aluminum present in the given mass by using the above unit formula as follows:
$\left( {\dfrac{{{\rm{1 \,mole\, Al}}}}{{{\rm{27\, g \,Al}}}}} \right) \times {\rm{8}}{\rm{.1\, g \,Al}} = {\rm{0}}{\rm{.3 \,mol\, Al}}$
We know that ${\rm{1 \,mole}}$ is the amount of substance that contains $6.02 \times {10^{23}}$ particles. Mathematically, we can express this relationship as per the following unit formula: $\dfrac{{6.02 \times {{10}^{23}}{\rm{ particles}}}}{{{\rm{1 \,mole}}}}$
For aluminum, particles are aluminum atoms so we can write the unit formula as follows:
$\dfrac{{6.02 \times {{10}^{23}}{\rm{ Al atoms}}}}{{{\rm{1 mole Al}}}}$
We can find out the number of atoms of aluminum present in the calculated number of moles by using the above unit formula as follows:
$\left( {\dfrac{{6.02 \times {{10}^{23}}{\rm{ Al\, atoms}}}}{{{\rm{1 \,mole \,Al}}}}} \right) \times {\rm{0}}{\rm{.3 \,mol\, Al}} = {\rm{1}}{\rm{.806}} \times {10^{23}}{\rm{\, Al \,atoms}}$
Now, we know that there are $4$ particles present in a single face-centred cubic unit cell. Mathematically, we can express the relationship between these two as per the following unit formula: $\dfrac{{{\rm{1 \,fcc \,unit\, cell}}}}{{4{\rm{ \,particles}}}}$
For aluminum, particles are aluminum atoms so we can rewrite the above unit formula for aluminum as follows:
$\dfrac{{{\rm{1\, fcc\, unit \,cell}}}}{{4{\rm{ \,Al \,atoms}}}}$
We can find out the number of f.c.c. unit cells of aluminum containing the calculated number of aluminum atoms by using the above unit formula as follows:
$\left( {\dfrac{{{\rm{1 \,fcc \,unit \,cell}}}}{{4{\rm{ Al\, atoms}}}}} \right) \times {\rm{1}}{\rm{.806}} \times {10^{23}}{\rm{ Al\, atoms}} = {\rm{4}}{\rm{.515}} \times {10^{22}}{\rm{ fcc\, unit\, cell}}$
Hence, the number of f.c.c. unit cells in ${\rm{8}}{\rm{.1 \,g}}$ of aluminum is ${\rm{4}}{\rm{.515}} \times {10^{22}}$.

Note:
For aluminum, particles are atoms but in other cases, it can be taken as per the nature of the substance, for example, formula units of sodium chloride.