A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g/cm3. Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of H2SO4 in this solution.

Mass of H₂SO₄ = 571.6 grams
Density of H₂SO₄ = 1.329 g/cm³ Density is the ratio of the weight (mass) of a substance to the volume it occupies. So, being 1 cm³=1 mL= 0.001 L, this value means that you have 1.329 grams per 1 mL or 1329 grams per 1 L.

Then:

a) Mass percentage

The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

[tex]Mass percentage=\frac{mass of solute}{mass of solution} x100[/tex]

In this case:

mass of solute= 571.6 grams
mass of solution= 1329 grams, taking into account 1 L of the solution

Replacing:

[tex]Mass percentage=\frac{571.6 grams}{1329 grams} x100[/tex]

Solving:

mass percentage= 43%

The mass percentage of the solution is 43%.

b) Mole fraction

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

Being the molar mass of H₂SO₄ equals to 98.08 [tex]\frac{g}{mol}[/tex], the number of moles of H₂SO₄ can be calculated as:

[tex]Number of moles H_{2} SO_{4} = \frac{mass H_{2} SO_{4}}{molar massH_{2} SO_{4}}[/tex]

[tex]Number of moles H_{2} SO_{4} = \frac{571.6 grams}{98.08\frac{g}{mol}}[/tex]

Number of moles H₂SO₄ = 5.83 moles

On the other side, being the mass of the solution 1329 grams in 1 L, the mass of water is:

mass of water= 1329 grams -571.6 grams= 757.4 grams

Then, being the molar mass of H₂O equals to 18.02 [tex]\frac{g}{mol}[/tex], the number of moles of H₂O can be calculated as:

[tex]Number of moles H_{2} O = \frac{mass H_{2} O}{molar massH_{2} O}[/tex]

[tex]Number of moles H_{2} O = \frac{757.4 grams}{18.02\frac{g}{mol}}[/tex]

Number of moles H₂O = 42.03 moles

So, the total moles of the solution can be calculated as:

Total moles = 5.83 + 42.03 = 47.86 moles

Finally, the more fraction can be calculated as follow:

[tex]Mole fraction H_{2} SO_{4} = \frac{moles of solute H_{2} SO_{4}}{total moles}[/tex]

[tex]Mole fraction H_{2} SO_{4} = \frac{5.83 moles}{47.86 moles}[/tex]

Mole fraction H₂SO₄ = 0.122

The mole fraction is 0.122

c) Molality

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{hilograms of solvent}[/tex]

In this case, you know:

number of moes of solute= 5.83 moles
Mass of solvent = 1329 grams - 571.6 grams = 757.4 grams = 0.7574 kg (being 1000 g=1 kg)

Replacing:

[tex]molality H_{2} SO_{4} = \frac{5.83 moles}{0.7574 kg}[/tex]

Molality H₂SO₄ = 7.70 [tex]\frac{moles}{kg}[/tex]

The molality of the solution is 7.70 [tex]\frac{moles}{kg}[/tex].

d) Molarity

Molarity is the number of moles of solute that are dissolved in a certain volume and is determined by the following expression:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

In this case, taking into account that the volume considered is 1 L, the molarity can be calculated as:

Then:

Molarity H₂SO₄ = 5.83 [tex]\frac{moles}{L}[/tex]

The molarity of the solution is 5.83 [tex]\frac{moles}{L}[/tex].

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