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Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is 0.824 g/cm^3 (cubed)

a.) Calculate the concentration of the solution in molality.
b.) Calculate the concentration of the solution in mole fraction of LiBr
c.) Calculate the concentration of the solution in mass percentage of CH3CN

Respuesta :

anfabba15

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

c. 81% by mass CH₃CN

Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, in this case / Total mass) . 100

We were asked for the acetonitrile → (667.7 g / 824 g) . 100 = 81%

mickymike92

Based on the calculations:

  • molality of the solution is 2.70 mol/kg
  • mole fraction of LiBr is 0.1
  • mass percentage of the solution is 81%

How to calculate the molality

The molality of a solution is calculated as follows:

  • Molality = moles of solute/ mass of solvent

Concentration of solute is 1.80 M

Thus; Moles of solute = 1.80 moles in 1 litre

Mass of solvent = Mass of solution - mass of solute

Mass of solution = volume × density

volume of solution= 1 L = 1000 cm^3

mass of solution = 1000 × 0.824

mass of solution = 824 g

Mass of solute = number of moles × molar mass

molar mass of solute = 86.84 g/mol

Mass of solute = 1.80 × 86.84

Mass of solute = 156.3 g

Mass of solvent = 824 - 156.3

Mass of solvent = 667.7 g = 0.6677 kg

molality of the solution = 1.80/0.6677

molality of the solution = 2.70 mol/kg

Mole fraction

  • Mole fraction = moles of solute/total moles
  • Total moles = moles of solute + moles of solvent

Moles of solvent = mass/molar mass

molar mass of solvent = 41 g/mol

Moles of solvent = 667.7/41

Moles of solvent = 16.3

Total moles = 16.3 + 1.8 = 18.1 moles

Moles fraction of LiBr = 1.8/18.1

Moles fraction of LiBr = 0.1

Percentage mass of Solvent

  • Percentage mass of Solvent = mass of solvent/mass of solution × 100%

Percentage mass of Solvent = 667.7/ 824 × 100 %

Percentage mass of Solvent = 81 %

Therefore, based on the calculations:

  • molality of the solution is 2.70 mol/kg
  • mole fraction of LiBr is 0.1
  • mass percentage of the solution is 81%

Learn more about molality, mole fraction and percentage mass of a substance at: https://brainly.com/question/14594475

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