Chemistry, 03.08.2019 04:00 morgaaaan651

The following table lists molar concentrations of seven major ions in seawater. using a density of 1.022 g/ml for seawater, convert the concentrations for the two ions in the question below into molality. ions g/kg mm na+ 10.781 480.57 k+ 0.399 10.46 mg2+ 1.284 54.14 ca2+ 0.4119 10.53 cl- 19.353 559.40 so42- 2.712 28.93 hco3- 0.126 2.11 total 35.067 n/a magnesium ion = m? sulfate ion= m?

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Answer from: sabahfayaskhan

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 :

$\rm Molality\;=\;moles\;of\;solute\;\times\;\dfrac{1000}{mass\;of\;solvent\;(g)}$

The mass of solvent in the solution is the mass of seawater without ions.

mass = density $\times$ volume

The density of seawater = 1.022 g/ml

The volume of seawater = 1000 ml

The mass of seawater = 1.022 g/ml $\times$ 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 $\times$ 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 $\rm \times\;10^-^3$ 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 $\rm \times\;10^-^3$ moles

Molality $\rm =\;moles\;of\;solute\;\times\;\dfrac{1000}{mass\;of\;solvent\;(g)}$

$\rm Molality\;of\;Mg^+\;=\;54.14\;\times\;10^-^3\;\dfrac{1000}{986.162}$

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 $\rm \times\;10^-^3$ M.

The moles of Sulfate ion in 1L of solution = 28.93 $\rm \times\;10^-^3$ moles.

$\rm Molality\;of\;SO_4^2^-\;=\;28.93\;\times\;10^-^3\;\dfrac{1000}{986.162}$

Molality of Sulfate ion = 0.0293 m.

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The following table lists molar concentrations of seven major ions in seawater. using a density of 1

Answer from: makayla912

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.

$\large{\boxed{\bold{m=n~\times~\frac{1000}{p} }}}$

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

$m = 54.10^{-3}.\frac{1000}{986.162}$

m = 0.0548

2. SO₄²⁻ ion

concentration = 28.93 mM = 28.93.10⁻³ M

mole SO₄²⁻ = 28.93.10⁻³

$m~=~28.93.10^{-3} \frac{1000}{986.192}$

m = 0.0293

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The following table lists molar concentrations of seven major ions in seawater. using a density of 1...