Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M
The correct answer to the question is Option C. 0.20 M Mg ion & 0.40 M Cl ion
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
The molarity of the diluted solution can be obtained as follow:
0.6 × 200 = M₂ × 600
120 = M₂ × 600
Divide both side by 600
M₂ = 120 / 600
Thus, the molarity of the diluted (i.e resulting) solution of MgCl₂ is 0.2 M
Next, we shall determine the concentration of magnesium ion, 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 magnesium ion, Mg²⁺ in the resulting solution is 0.2 M.
Finally, we shall determine the concentration of the chloride ion, Cl¯ in the resulting solution.
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole MgCl₂ dissolves to produce 2 moles of Cl¯
Therefore,
0.2 M MgCl₂ will also produce = 2 × 0.2 = 0.4 M Cl¯
Thus, the concentration of chloride ion, Cl¯ in the resulting solution is 0.4 M.
From the calculations made above:
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M
Option C. 0.20 M Mg ion & 0.40 M Cl ion gives the correct answer to the question.
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