Answer:
M of MgCl₂ = 1.65 × 10⁻⁶ M
M of Mg²⁺ = 1.65 × 10⁻⁶ M
M of Cl⁻ = 3.30 × 10⁻⁶ M
Explanation:
1) MgCl₂
Molarity = number of moles of solute / volume of solution in liters, M = n / V
n = mass in grams / molar mass
molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol
n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles
⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M
2) Mg²⁺ and Cl⁻
Those are the ions in solution.
You assume 100% dissociation of the ionic compound (strong electrolyte).
Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻
That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.
That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.
So, the results are:
M of MgCl₂ = 1.65 × 10⁻⁶ M
M of Mg²⁺ = 1.65 × 10⁻⁶ M
M of Cl⁻ = 3.30 × 10⁻⁶ M
