Question

A sample of $Al{F_3}$ contains $3.0 \times {10^{24}}$ ${F^ - }$ ions. The number of formula units of the sample are:
A. $9.0 \times {10^{24}}$
B. $3.0 \times {10^{24}}$
C. $0.75 \times {10^{24}}$
D. $1.0 \times {10^{24}}$

Answer 1

The mole is the unit of measurement for amount of substance in the International System of Units. A mole of a substance or a mole of particles is defined as exactly particles, which may be atoms, molecules, ions, or electrons. In short, for particles, .

Complete answer:
Aluminium fluoride refers to inorganic compounds with the formula$Al{F_3}\cdot x{H_2}O$ . They are all colorless solids. Anhydrous $AlF_3$ is used in the production of aluminium metal. Several occur as minerals.
In the given question, it has been given that a sample of $Al{F_3}$ contains $3.0 \times {10^{24}}$ ${F^ - }$ ions. The dissociation of one aluminium fluoride molecule can be shown from the following reaction:
$Al{F_3} \to A{l^{3 + }} + 3{F^ - }$
Thus, we can say that one mole of aluminium fluoride ($Al{F_3}$) produces three moles of fluoride (${F^ - }$) ions.
Let the number of formula units of aluminium fluoride be $x$ .
One mole of $Al{F_3}$dissociates to yield = 3 moles of fluoride ions.
$x$ moles of $Al{F_3}$dissociates to yield = $3x$ moles of fluoride ions.
One mole consists of = $6 \times {10^{23}}ions$
$3 \times 6 \times {10^{23}}$ ions of fluorine is produced by = $6 \times {10^{23}}$ ions of $Al{F_3}$
$1$ ion of fluorine is produced by = $\dfrac{{6 \times {{10}^{23}}}}{{3 \times 6 \times {{10}^{23}}}}$ ions of $Al{F_3}$
$3.0 \times {10^{24}}$ ${F^ - }$ ions are produced by = $\dfrac{{6 \times {{10}^{23}}}}{{3 \times 6 \times {{10}^{23}}}} \times 3 \times {10^{24}} = 1 \times {10^{24}}$ ions of $Al{F_3}$.
Thus, the number of formula units of the sample is $1.0 \times {10^{24}}$ .

Thus option D is the correct answer.

Note:
Aluminium fluoride is an important additive for the production of aluminium by electrolysis. Together with cryolite, it lowers the melting point to below $1000\;^\circ C$ and increases the conductivity of the solution. It is into this molten salt that aluminium oxide is dissolved and then electrolyzed to give bulk $Al$ metal.