Respuesta :
Answer:
Final concentration of Mg2+ = 0.20 M
Explanation:
Concentration of [tex]MgCl_2(aq)\ (M_1)[/tex] = 0.60 M
Volume of [tex]MgCl_2(aq)\ (V_1)[/tex] = 200 mL
Volume of distilled water added = 400 mL
Final volume of the soution = 200 mL + 400 mL
= 600 mL
Final concentration of solution = [tex]M_2[/tex]
The final concentration is calculated as follows:
[tex]M_1 V_1=M_2V_2\\0.60 \times 200= M_2 \times 600\\M_2=\frac{0.60\times 200}{600} =0.20\ M[/tex]
Therefore, final concentration of the solution is 0.20 M.
[tex]MgCl_2(aq)[/tex] exists in the solution as Mg2+ and 2Cl-.
Therefore, concentration of Mg2+ is same as the final concentration of solution.
Final concentration of Mg2+ = 0.20 M
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M.
To solve this question, we'll begin by calculating the molarity of the diluted solution. This can be obtained as follow:
Volume of stock solution (V₁) = 200 mL
Molarity of stock solution (M₁) = 0.60 M
Volume of diluted solution (V₂) = 200 + 400 = 600 mL
Molarity of diluted solution (M₂) = ?
The molarity of the diluted solution can be obtained as follow:
M₁V₁ = M₂V₂
0.6 × 200 = M₂ × 600
120 = M₂ × 600
Divide both side by 600
M₂ = 120 / 600
M₂ = 0.2 M
Thus, the molarity of the diluted solution of MgCl₂ is 0.2 M
Finally, we shall determine the concentration of Mg²⁺ in the diluted solution. This is illustrated below:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole MgCl₂ dissolves to produce 1 mole Mg²⁺.
Therefore,
0.2 M MgCl₂ will also produce 0.2 M Mg²⁺.
Thus, the concentration of Mg²⁺ in the resulting solution is 0.2 M.
Learn more: https://brainly.com/question/24848714
