Answer:
Approximately [tex]19.1\; \rm g[/tex].
Explanation:
Number of moles of formula units of magnesium sulfate required to make the solution
The unit of concentration in this question is "[tex]\rm M[/tex]". That's equivalent to "[tex]\rm mol \cdot L^{-1}[/tex]" (moles per liter.) In other words:
[tex]c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}[/tex].
However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:
[tex]\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L[/tex].
Calculate the number of moles of [tex]\rm MgSO_4[/tex] formula units required to make this solution:
[tex]\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}[/tex].
Mass of magnesium sulfate in the solution
Look up the relative atomic mass data 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]\rm 32.06[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
Calculate the formula mass of [tex]\rm MgSO_4[/tex] using these values:
[tex]M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}[/tex].
Using this formula mass, calculate the mass of that (approximately) [tex]0.159\; \rm mol[/tex] of [tex]\rm MgSO_4[/tex] formula units:
[tex]\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}[/tex].
Therefore, the mass of [tex]\rm MgSO_4[/tex] required to make this solution would be approximately [tex]19.1\; \rm g[/tex].
