Question

Determine the concentrations of MgCl2, Mg2+, and Cl− in a solution prepared by dissolving 2.52 × 10−4 g MgCl2 in 2.00 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm).

Answer 1

Step-by-step explanation:

Given parameters:

Mass of MgCl₂ = 2.52 x 10⁻⁴g

Volume of water = 2L

Unkown:

Concentration of MgCl₂ =?

Concentration of Mg²⁺ = ?

Concentration of Cl⁻ =?

Solution:

Concentration is defined as the number of moles of a solute contained in a solution.

Concentration =
`(number of moles )/(volume)`

To find the number of moles"

number of moles =
`(mass)/(molar mass)`

Molar mass of MgCl₂ = 24.3 + (2 x 35.5) = 95.3g/mol

number of moles =
`(0.000)/(molar mass)` = 0.000252moles

Concentration of MgCl₂ =
`(0.000252)/(95.3)` = 2.64 x 10⁻⁶moldm⁻³

from the formula of the compound;

1 mole of MgCl₂ contains 1 mole of Mg²⁺

Therefore, 2.64 x 10⁻⁶moldm⁻³ of MgCl₂ will contains 2.64 x 10⁻⁶moldm⁻³ of Mg

Also;

1 mole of MgCl₂ contains 2 mole of Cl⁻

2.64 x 10⁻⁶moldm⁻³ contains 2 x 2.64 x 10⁻⁶moldm⁻³; 5.29 x 10⁻⁶moldm⁻³

Expressing in ppm;

1ppm = 1mg/L

2.64 x 10⁻⁶moldm⁻³ to mg/L for Mg²⁺

2.64 x 10⁻⁶moldm⁻³ = 2.64 x 10⁻⁶moldm⁻³ x molar mass(g/mol)

= 2.64 x 10⁻⁶moldm⁻³ x 24.3 = 6.43 x 10⁻⁵g/L

g/L to mg/L; 6.43 x 10⁻⁵g/L x 1000 = 6.43 x 10⁻²mg/L = 6.43 x 10⁻²ppm

5.29 x 10⁻⁶moldm⁻³ to mg/L for Cl⁻;

5.29 x 10⁻⁶moldm⁻³ = 5.29 x 10⁻⁶moldm⁻³ x 35.5 = 1.88 x 10⁻⁴g/L

g/L to mg/L; 1.88 x 10⁻⁴g/L x 1000 = 1.88 x 10⁻¹mg/L = 1.88 x 10⁻¹ppm