Answer:
<em>3.88x10⁻²</em>
Explanation:
First, the amount of moles initially in the volume of the concentrated MgCl2 solution is determined:
1000 mL solution ____ 2.19 moles of MgCl₂
27.00 mL solution _____ X = 5.91x10⁻² mol of MgCl₂
<em>Calculation:</em> 27.00 mL x 2.19 moles / 1000 mL = 0.05913 moles ≡ 5.91x10⁻² mol of MgCl₂
This is the amount of moles of MgCl₂ in both the initial and the diluted solution, since only water was added to the first solution, therefore, in 122.00 mL of the new solution, there are 5.91x10⁻² mol of MgCl₂. Now it is necessary to calculate the amount of moles in 40 mL of the new solution prepared:
122.00 mL of solution _____ 5.91x10⁻² mol of MgCl₂
40.00 mL of solution ____ X = 1.94x10⁻² of MgCl₂
<em>Calculation:</em> 40.00 mL x 5.91x10⁻² mol / 122.00 mL = 0.01939 moles ≡ 1.94x10⁻² MgCl₂
Now, when the MgCl₂ compound is in an aqueous solution, it dissociates into the ions that form it as follows:
MgCl₂ ⇒ Mg⁺² (aq) + 2 Cl⁻ (aq)
Therefore, for each mole of MgCl₂ that dissolves in water, it will dissociate producing two moles of Cl⁻ chloride ion. Mathematically:
1 mole of MgCl₂ ____ 2 moles of Cl⁻
1.94x10⁻² moles of MgCl₂ _____ X = 3.88x10⁻² moles of Cl⁻
<em>Calculation:</em> 1.94x10⁻² moles x 2 moles / 1 mole = 0.03877 ≡ 3.88x10⁻² moles of Cl⁻
Therefore, <em>in 40.00 mL of the new solution, there will be a concentration of 3.88x10⁻² moles of Cl⁻</em>
