To solve the problem of finding the mass of solid MgCO₃ formed when a 500 mL saturated solution is reduced to 120 mL by evaporation, we need to follow several steps. Here is the detailed, step-by-step solution:
1. Calculate the molar solubility of MgCO₃:
The solubility product constant (Ksp) for MgCO₃ is given as [tex]\(4.0 \times 10^{-5}\)[/tex].
The dissociation equation of MgCO₃ is:
[tex]\[
\text{MgCO}_3 \rightleftharpoons \text{Mg}^{2+} + \text{CO}_3^{2-}
\][/tex]
If the molar solubility of MgCO₃ is [tex]\(S\)[/tex], then:
[tex]\[
K_{sp} = [\text{Mg}^{2+}][\text{CO}_3^{2-}] = S \times S = S^2
\][/tex]
Solving for [tex]\(S\)[/tex]:
[tex]\[
S = \sqrt{K_{sp}} = \sqrt{4.0 \times 10^{-5}} = 0.00632 \text{ mol/L}
\][/tex]
2. Calculate the amount of MgCO₃ in the initial solution:
The initial volume of the solution is 500 mL, which is 0.5 L (since 1000 mL = 1 L).
The number of moles of MgCO₃ in 0.5 L of solution:
[tex]\[
\text{moles} = \text{molar solubility} \times \text{volume (in L)} = 0.00632 \times 0.5 = 0.00316 \text{ moles}
\][/tex]
The mass of MgCO₃ in grams:
[tex]\[
\text{mass} = \text{moles} \times \text{molar mass} = 0.00316 \times 84 = 0.26563 \text{ g}
\][/tex]
3. Calculate the amount of MgCO₃ in the final solution:
The final volume of the solution is 120 mL, which is 0.12 L.
The number of moles of MgCO₃ in 0.12 L of solution:
[tex]\[
\text{moles} = \text{molar solubility} \times \text{volume (in L)} = 0.00632 \times 0.12 = 0.00076 \text{ moles}
\][/tex]
The mass of MgCO₃ in grams:
[tex]\[
\text{mass} = \text{moles} \times \text{molar mass} = 0.00076 \times 84 = 0.06375 \text{ g}
\][/tex]
4. Calculate the mass of MgCO₃ formed:
The mass of MgCO₃ formed by evaporation will be the difference between the mass of MgCO₃ in the initial and final solutions:
[tex]\[
\text{mass formed} = \text{mass}_{\text{initial}} - \text{mass}_{\text{final}} = 0.26563 - 0.06375 = 0.20188 \text{ g}
\][/tex]
Therefore, the mass of solid MgCO₃ that is formed is approximately 0.20 g.
The correct answer is:
(C) 0.20 g
