Question

What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M solution?

Answer 1

Answer:

Approximately $19.1\; \rm g$.

Explanation:

Number of moles of formula units of magnesium sulfate required to make the solution

The unit of concentration in this question is "$\rm M$". That's equivalent to "$\rm mol \cdot L^{-1}$" (moles per liter.) In other words:

$c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}$.

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

$\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L$.

Calculate the number of moles of $\rm MgSO_4$ formula units required to make this solution:

$\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}$.

Mass of magnesium sulfate in the solution

Look up the relative atomic mass data of $\rm Mg$, $\rm S$, and $\rm O$ on a modern periodic table:

• $\rm Mg$: $24.305$.
• $\rm S$: $\rm 32.06$.
• $\rm O$: $15.999$.

Calculate the formula mass of $\rm MgSO_4$ using these values:

$M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}$.

Using this formula mass, calculate the mass of that (approximately) $0.159\; \rm mol$ of $\rm MgSO_4$ formula units:

$\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}$.

Therefore, the mass of $\rm MgSO_4$ required to make this solution would be approximately $19.1\; \rm g$.