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
Answer 1) : To calculate the concentration of number of moles of Mg[tex] Cl_{2} [/tex], by dissolving 2.75 X [tex] 10^{-4} [/tex] of Mg[tex] Cl_{2} [/tex] will be like this;
Number of moles of Mg[tex] Cl_{2} [/tex] = (2.75 X [tex] 10^{-4} [/tex] g of Mg[tex] Cl_{2} [/tex]) X (1 mole of Mg[tex] Cl_{2} [/tex] / 95.211 g of Mg[tex] Cl_{2} [/tex])
= 2.88 X [tex] 10^{-6} [/tex] moles of Mg[tex] Cl_{2} [/tex] (g/L)
In ppm it will be 0.288 ppm
Answer 2) To find the moles of Mg ions in solution of Mg[tex] Cl_{2} [/tex], we need to find the moles of Mg present in the solution,
Here, moles of Mg = Moles of Mg[tex] Cl_{2} [/tex]
So, 2.88 X [tex] 10^{-6} [/tex] moles of Mg[tex] Cl_{2} [/tex] = 2.88 X [tex] 10^{-6} [/tex] moles of Mg. (g/L)
And in ppm it will be 0.288 ppm
Answer 3) For calculating the moles of Cl ions present in the solution of Mg[tex] Cl_{2} [/tex], we need to know how many moles are present in Mg[tex] Cl_{2} [/tex],
We see 2 moles of Cl ions are present in solution of Mg[tex] Cl_{2} [/tex]
Therefore, 2 moles of Cl = 1 mole of Mg[tex] Cl_{2} [/tex]
So, We can multiply 2 with the molarity of Mg[tex] Cl_{2} [/tex], we get,
2 X (2.88 X [tex]10^{-6}[/tex])= 5.77 X [tex]10^{-6}[/tex] g/L
And in ppm it will be 0.577 ppm.
