Answer:
Approximately [tex]5.65 \times 10^{-2}\; \rm mol[/tex], which is approximately [tex]3.41 \times 10^{22}[/tex] particles.
Explanation:
Formula of magnesium sulfite: [tex]\rm MgSO_3[/tex].
Look up the relative atomic mass of [tex]\rm Mg[/tex]. [tex]\rm S[/tex], and [tex]\rm O[/tex] on a modern periodic table:
- [tex]\rm Mg[/tex]: [tex]24.305[/tex].
- [tex]\rm S[/tex]: [tex]32.06[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
The ionic compound [tex]\rm MgSO_3[/tex] consist of magnesium ions [tex]\rm {Mg}^{2+}[/tex] and sulfite ions [tex]\rm {SO_3}^{2-}[/tex].
Notice that [tex]\rm {Mg}^{2+}\![/tex] and [tex]\rm {SO_3}^{2-}\![/tex] combine at a one-to-one ratio to form the neural compound [tex]\rm MgSO_3[/tex]. Therefore, each [tex]\rm MgSO_3\![/tex] formula unit would include one [tex]\rm {Mg}^{2+}[/tex] ion and one [tex]\rm {SO_3}^{2-}[/tex] ion (that would be two ions in total).
Calculate the formula mass of one such formula unit:
[tex]\begin{aligned}&\; M(\mathrm{MgSO_3}) \\ = & \; 24.305 + 32.06 + 3 \times 15.999 \\ = & \; 104.362\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
In other words, the mass of one mole of [tex]\rm MgSO_3[/tex] formula units (which includes one mole of [tex]\rm {Mg}^{2+}[/tex] ions and one mole of [tex]\rm {SO_3}^{2-}[/tex] ions) would be [tex]104.362\; \rm g[/tex].
Calculate the number of moles of such formula units in that [tex]2.95\; \rm g[/tex] of [tex]\rm MgSO_3[/tex]:
[tex]\begin{aligned}n&= \frac{m}{M}\\ &=\frac{2.95\; \rm g}{104.362\; \rm g \cdot mol^{-1}} \approx 2.82670 \times 10^{-2}\; \rm mol\end{aligned}[/tex].
There are two moles of ions in each mole of [tex]\rm MgSO_3[/tex] formula units. Therefore, that [tex]2.82670 \times 10^{-2}\; \rm mol[/tex] of [tex]\rm MgSO_3\![/tex] formula units would include approximately [tex]2 \times 2.82670 \times 10^{-2}\; \rm mol \approx 5.65\times 10^{-2}\; \rm mol[/tex] of ions.
