Question

Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water. express all three concentrations in molarity. additionally, express the concentrations of the ionic species in parts per million (ppm).

Answer 1

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 2

Answer 1) : To calculate the concentration of number of moles of Mg$Cl_{2}$, by dissolving 2.75 X $10^{-4}$ of Mg$Cl_{2}$ will be like this;

Number of moles of Mg$Cl_{2}$ = (2.75 X $10^{-4}$ g of Mg$Cl_{2}$) X (1 mole of Mg$Cl_{2}$ / 95.211 g of Mg$Cl_{2}$)

= 2.88 X $10^{-6}$ moles of Mg$Cl_{2}$ (g/L)

In ppm it will be 0.288 ppm

Answer 2) To find the moles of Mg ions in solution of Mg$Cl_{2}$, we need to find the moles of Mg present in the solution,

Here, moles of Mg = Moles of Mg$Cl_{2}$

So, 2.88 X $10^{-6}$ moles of Mg$Cl_{2}$ = 2.88 X $10^{-6}$ 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$Cl_{2}$, we need to know how many moles are present in Mg$Cl_{2}$,

We see 2 moles of Cl ions are present in solution of Mg$Cl_{2}$

Therefore, 2 moles of Cl = 1 mole of Mg$Cl_{2}$

So, We can multiply 2 with the molarity of Mg$Cl_{2}$, we get,

2 X (2.88 X $10^{-6}$)= 5.77 X $10^{-6}$ g/L

And in ppm it will be 0.577 ppm.