If 200. ml of 0.60 m mgcl2(aq) is added to 400. ml of distilled water, what is the concentration of mg2+(aq) in the resulting solution? (assume volumes are additive).

Final concentration of Mg2+ = 0.20 M

Explanation:

Concentration of = 0.60 M

Volume of = 200 mL

Volume of distilled water added = 400 mL

Final volume of the soution = 200 mL + 400 mL

                                              = 600 mL

Final concentration of solution =

The final concentration is calculated as follows:

Therefore, final concentration of the solution is 0.20 M.

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

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

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.

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the answer is c i believe

i think its 5049.

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