Answer:
Approximately [tex]1.53 \times 10^{23}[/tex] formula units ([tex]0.254\; \rm mol[/tex]).
Explanation:
Refer to a modern periodic table for the relative atomic mass of magnesium ([tex]\rm Mg[/tex]) and chlorine ([tex]\rm Cl[/tex]):
In other words, the mass of [tex]1\; \rm mol[/tex] of [tex]\rm Mg[/tex] atoms would be (approximately) [tex]24.305\; \rm g[/tex].
Likewise, the mass of [tex]1\; \rm mol[/tex] of [tex]\rm Cl[/tex] atoms would be approximately [tex]35.45\; \rm g[/tex].
One formula unit of the ionic compound [tex]\rm MgCl_{2}[/tex] includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of [tex]1\; \rm mol[/tex] of the formula units of this compound.
The formula [tex]\rm MgCl_{2}[/tex] includes one [tex]\rm Mg[/tex] atom and two [tex]\rm Cl[/tex] atoms.
Hence, every formula unit of [tex]\rm MgCl_{2} \![/tex] would include the same number of atoms: one [tex]\rm Mg\![/tex] atom and two [tex]\rm Cl\![/tex] atoms. There would be [tex]1\; \rm mol[/tex] of [tex]\rm Mg[/tex] atoms and [tex]2\; \rm mol[/tex] of [tex]\rm Cl[/tex] atoms in [tex]1\; \rm mol\![/tex] of [tex]\rm MgCl_{2}[/tex] formula units.
Thus, the mass of [tex]1\; \rm mol\![/tex] of [tex]\rm MgCl_{2}[/tex] formula units would be equal to the mass of [tex]1\; \rm mol[/tex] of [tex]\rm Mg[/tex] atoms plus the mass of [tex]2\; \rm mol[/tex] of [tex]\rm Cl[/tex] atoms. (The mass of [tex]1\; \rm mol\!\![/tex] of each atom could be found from the relative atomic mass of each element.)
[tex]\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
In other words, the formula mass of [tex]\rm MgCl_{2}[/tex] is [tex]95.205\; \rm g \cdot mol^{-1}[/tex].
Therefore, the number of formula units in [tex]m = 24.2\; \rm g[/tex] of [tex]\rm MgCl_{2}[/tex] would be:
[tex]\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}[/tex].
Multiple [tex]n[/tex] by Avogadro's Number [tex]N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}[/tex] to estimate the number of formula units in [tex]0.254\; \rm mol[/tex]:
[tex]\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}[/tex].
