Respuesta :
Molality of Mg²⁺ ions = 0.0548
Molality of SO₄²⁻ ions = 0.0293
Further explanation
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. Concentration shows the amount of solute in a unit of the amount of solvent.
Molality (m)
Molality shows how many moles are dissolved in every 1000 grams of solvent.
[tex]\large{\boxed{\bold{m=n~\times~\frac{1000}{p} }}}[/tex]
m = Molality
n = number of moles of solute
p = solvent mass (grams)
To find the molality of these ions we determine the value of solvents and dissolved moles (moles of magnesium ions and sulfate ions)
We use the calculation base of 1 Liter of solution so that the mass of the seawater.
seawater mass: volume x density
seawater mass: 1000 ml (1 L) x 1,022 g / mL
seawater mass: 1022 grams = 1,022 kg
then the total mass of ions in solution =
35,067 gr / kg x 1,022 kg = 35,838 gr
Solvent mass = seawater mass - ion mass
mass of the solvent = 1022 - 35,838
solvent mass = 986,162 grams
- 1. Mg²⁺ ions
concentration = 54.14 mM = 54.10⁻³ M
because it is counted in 1 liter of solution then Mg²⁺ = 54.10⁻³ mole ion
[tex]m = 54.10^{-3}.\frac{1000}{986.162}[/tex]
m = 0.0548
- 2. SO₄²⁻ ion
concentration = 28.93 mM = 28.93.10⁻³ M
mole SO₄²⁻ = 28.93.10⁻³
[tex]m~=~28.93.10^{-3} \frac{1000}{986.192}[/tex]
m = 0.0293
Learn more
The molarity of luminol
https://brainly.com/question/2814870
moles of water you can produce
https://brainly.com/question/1405182
moles of NaOH
https://brainly.com/question/4283309
Keywords: molal, seawater, density
The seawater has the Mg ion molality to be 0.0548 m. The molality of Sulfate ion on the seawater is 0.0293 m.
Molality is used to define the concentration of the solute in the 1000 grams of solvent.
The molality can be calculated as :
[tex]\rm Molality\;=\;moles\;of\;solute\;\times\;\dfrac{1000}{mass\;of\;solvent\;(g)}[/tex]
The mass of solvent in the solution is the mass of seawater without ions.
mass = density [tex]\times[/tex] volume
The density of seawater = 1.022 g/ml
The volume of seawater = 1000 ml
The mass of seawater = 1.022 g/ml [tex]\times[/tex] 1000 ml
The mass of seawater = 1022 g.
Given the total mass of ions in the seawater = 35.067 g/kg
i.e., in 1 kg of seawater 35.067 g of ions are present.
The mass of ions in 1022g (1.022 kg) of seawater = 1.022 [tex]\times[/tex] 35.067 g
The mass of ions in the seawater = 35.838 grams.
The mass of solvent = Total mass of seawater - the mass of ions
The mass of solvent = 1022 grams - 35.838 grams
The mass of solvent = 986.162 grams
- The molality of magnesium ion:
The concentration of Mg ion = 54.15 mM
The concentration of Mg ion = 54.15 [tex]\rm \times\;10^-^3[/tex] M.
Molarity is the moles of solvent in a liter of solution.
Since the volume of the solution is assumed to be 1 L.
The moles of Mg ion in 1L of solution = 54.15 [tex]\rm \times\;10^-^3[/tex] moles
Molality [tex]\rm =\;moles\;of\;solute\;\times\;\dfrac{1000}{mass\;of\;solvent\;(g)}[/tex]
[tex]\rm Molality\;of\;Mg^+\;=\;54.14\;\times\;10^-^3\;\dfrac{1000}{986.162}[/tex]
Molality of Mg ion = 0.0548 m.
- The molality of Sulfate ion:
The concentration of Sulfate ion = 28.93 mM
The concentration of Sulfate ion = 28.93 [tex]\rm \times\;10^-^3[/tex] M.
The moles of Sulfate ion in 1L of solution = 28.93 [tex]\rm \times\;10^-^3[/tex] moles.
[tex]\rm Molality\;of\;SO_4^2^-\;=\;28.93\;\times\;10^-^3\;\dfrac{1000}{986.162}[/tex]
Molality of Sulfate ion = 0.0293 m.
For more information, refer to the link:
https://brainly.com/question/23475324
