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Automobile battery acid is 38% H2SO4 and has a destiny of 1.29g/ml. Calculate the molality and the molarity of this solution. ​

Respuesta :

sebassandin

Answer:

[tex]M=5.0M\\\\m=6.2m[/tex]

Explanation:

Hello,

In this case, 38 % is commonly a by mass concentration, meaning that we have 38 grams of solute (sulfuric acid) per 100 grams of solution (water+sulfuric acid):

[tex]38\%=\frac{m_{H_2SO_4}}{m_{H_2SO_4}+m_{H_2O}}[/tex]

Hence, we compute the moles of sulfuric acid in 38 grams by using its molar mass (98 g/mol):

[tex]n_{H_2SO_4}=38g*\frac{1mol}{98g}=0.39mol H_2SO_4[/tex]

Next, the volume of the solution in litres by using the density of the solution:

[tex]V_{solution}=100g*\frac{1mL}{1.29g}*\frac{1L}{1000mL} =0.0775L[/tex]

This is done since the molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, thus we have:

[tex]M=\frac{n}{V}=\frac{0.38mol}{0.0775L}=5.0M[/tex]

On the other hand, the molality is defined as the ratio of the moles of the solute to the mass of the solvent in kilograms, thus, we compute the mass of water (solvent) as shown below:

[tex]m_{H_2O}=100g-38g=62g*\frac{1kg}{1000g}=0.062kg[/tex]

So compute the molality:

[tex]m=\frac{n_{solute}}{m_{solvent}}=\frac{0.39mol}{0.062kg}=6.2m[/tex]

Regards.

Eduard22sly

1. The molarity of the solution is 5 M

2. The molality of the solution is 6.26 M

Let the mass of the solution be 100 g.

Therefore, the mass of 38% of H₂SO₄ in the solution is 38 g.

Next, we shall determine the mole of 38 g of H₂SO₄.

  • Mass of H₂SO₄ = 38 g
  • Molar mass of H₂SO₄ = (2×1) + 32 + (16×4) = 98 g/mol
  • Mole of H₂SO₄ =?

Mole = mass / molar mass

Mole of H₂SO₄ = 38 / 98

Mole of H₂SO₄ = 0.388 mole

Next, we shall determine the volume of the solution

  • Mass of solution = 100 g
  • Density of solution = 1.29 g/mL

Volume of solution =?

Volume = mass / density

Volume of solution = 100 / 1.29

Volume of solution = 77.52 mL

1. Determination of the molarity of the solution

  • Mole of H₂SO₄ = 0.388 mole
  • Volume of solution = 77.52 mL = 77.52/1000 = 0.07752 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.388 / 0.07752

Molarity = 5 M

2. Determination of the molality of the solution

  • Mole of H₂SO₄ = 0.388 mole
  • Mass of H₂SO₄ = 38 g
  • Mass of solution = 100 g
  • Mass of water = 100 – 38 = 62 g
  • Mass of water = 62 / 1000 = 0.062 Kg
  • Molality =?

Molality = mole / mass (Kg) of water

Molality = 0.388 / 0.062

Molality = 6.26 M

Learn more about concentration of solution:

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