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What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water? What mass of water is needed to prepare a 1.20 molal solution using 0.60 mol propyleneglycol? What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?

Respuesta :

Answer:

1) The molality is 0.145 molal

2) We need 0.50 kg water

3) The molality is 0.245 molal

Explanation:

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water?

Molality = moles AlCl3 / mass water (in kg)

Molality = 0.32 moles / 2.2 kg water

Molality = 0.145 moles/kg = 0.145 molal

The molality is 0.145 molal

What mass of water is needed to prepare a 1.20 molal solution using 0.60 mol propyleneglycol?

Molality = moles propyleneglycol / mass water

1.20 = 0.60 / x kg

X = 0.60 / 1.2

X = 0.50 kg water

We need 0.50 kg water

What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?

Molality = moles CO2 / mass water (in kg)

Molality = 0.145 moles / 0.591 kg

Molality = 0.245 molal

The molality is 0.245 molal
  • The molality of a solution in which 0.32 moles AlCl₃ has been dissolved in 2,200g water is 0.145 molal
  • The mass of water needed to prepare 1.20 molal solution using 0.60 mol propyleneglycol is 500 g
  • The molality of the solution prepared from 0.145 mole of  CO₂ and 591 g of water is 0.245 molal

Concentration in molality

Molality is the number of moles of solute per kilogram of solvent.

[tex]Molarity = \frac{Number\ of \ moles\ of \ solute}{Kg \ of\ solvent}[/tex]

  • For the molality of the AlCl₃ solution

Number of moles = 0.32 mole

Mass of solvent = 2200 g = 2.2 kg

Therefore,

Molality of the AlCl₃ solution = [tex]\frac{0.32}{2.2}[/tex]

Molality of the AlCl₃ solution = 0.145 molal

Hence, the molality of a solution in which 0.32 moles AlCl₃ has been dissolved in 2,200g water is 0.145 molal

  • For the mass of water needed to prepare 1.20 molal solution of the propyleneglycol

Number of moles of propyleneglycol = 0.60 mol

Molality = 1.20 molal

Therefore,

Mass of water (solvent) = [tex]\frac{0.60}{1.20}[/tex]

Mass of water = 0.5 kg

Mass of water = 500 g

Hence, the mass of water needed to prepare 1.20 molal solution using 0.60 mol propyleneglycol is 500 g

  • For the molality of the CO₂ solution

Number of moles of CO₂ = 0.145 mol

Mass of the solvent (water) = 591 g = 0.591 kg

Therefore,

Molality of the CO₂ solution = [tex]\frac{0.145}{0.591}[/tex]

Molality of the CO₂ solution = 0.245 molal

Hence, the molality of the solution prepared from 0.145 mole of  CO₂ and 591 g of water is 0.245 molal

Learn more on Concentration in molality here: https://brainly.com/question/4580605

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