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

Answer:

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 = $\frac{number of moles }{volume}$

 To find the number of moles"

           number of moles = $\frac{mass}{molar mass}$

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

       number of moles = $\frac{0.000}{molar mass}$ = 0.000252moles

 Concentration of MgCl₂ = $\frac{0.000252}{95.3}$ = 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