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Common commercial acids and bases are aqueous solutions with the following properties: Density Mass Percent of Solute Hydrochloric acid 1.19 38 Nitric acid 1.42 70. Sulfuric acid 1.84 95 Acetic acid 1.05 99 Ammonia 0.90 28 Calculate the molarity, molality, and mole fraction of each of the preceding reagents?

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Answer:

[HCl] : 12.3 M ; [HCl] : 16.7 m; [HCl] in mole fraction = 0.23

[HNO₃]: 15.7 M; [HNO₃]: 30 m ; [HNO₃] in mole fraction = 0.39

[H₂SO₄]: 17.8 M ; [H₂SO₄]: 193.8 m ; [H₂SO₄] in mole fraction = 0.78

[CH₃COOH]: 17.3 M ; [CH₃COOH]: 1650 m ; [CH₃COOH] in mole fraction: 0.96

[NH₃]: 14.7 M ; [NH₃]: 22.7 m ; [NH₃] in mole fraction = 0.29

Explanation:

Percent by mass means X g of solute contained in 100 g of solution

Mass of solvent: 100 g - X % by mass

Solution density = Solution mass / Solution volume

Solution Volume = Solution mass / Solution density

Molality = mol of solute / kg of solvent

Molarity = mol of solute / L of solution

Mole fraction = Moles of solute / Total moles

Total moles = moles of solute + moles of solvent

  • HCl → 1.19 g/mL; 38%

Mass of solvent: 100 g - 38 g = 62 g

Moles of solvent → 62 g . 1 mol/ 18g = 3.44 moles

Mass of solute → 38 g ; Moles of solute → 38 g . 1mol/36.45 g = 1.04 moles

Total moles = 3.44 mol + 1.04 mol = 4.48 moles

Mole fraction = 1.04 / 4.48 = 0.23

Mass of solvent → from g to kg → 62 g . 1kg/1000g = 0.062kg

Molality → 1.04 mol / 0.062 kg = 16.7 mol/kg → 16.7 m

Solution volume = 100 g / 1.19 g/mL → 84.03 mL

Solution volume → from mL to L → 84.03 mL . 1L/ 1000mL = 0.0840L

Molarity → 1.04 mol / 0.0840L = 12.3 M

HNO₃ → 1.42 g/mL; 70%

Mass of solvent: 100 g - 70 g = 30 g

Moles of solvent → 30 g . 1 mol/ 18g = 1.67  moles

Mass of solute → 70 g ; Moles of solute → 70 g . 1mol/ 63 g = 1.11 moles

Total moles = 1.67 mol + 1.11 mol = 2.78 moles

Mole fraction = 1.11 / 2.78 = 0.39

Mass of solvent → from g to kg → 30 g . 1kg/1000g = 0.030kg

Molality → 1.11 mol / 0.030 kg → 30 m

Solution volume = 100 g / 1.42 g/mL → 70.4mL

Solution volume → from mL to L → 70.4 mL . 1L/ 1000mL = 0.0704L

Molarity → 1.11 mol / 0.0704L = 15.7 M

  • H₂SO₄ →  1.84 g/mL;  95 %

Mass of solvent: 100 g - 95 g = 5 g

Moles of solvent → 5 g . 1 mol/ 18g = 0.277  moles

Mass of solute → 95 g ; Moles of solute → 95 g . 1 mol/ 98 g = 0.969 moles

Total moles = 0.277 mol + 0.969 mol = 1.246 moles

Mole fraction = 0.969 / 1.246 = 0.78

Mass of solvent → from g to kg → 5 g . 1kg/1000g = 0.005 kg

Molality → 0.969 mol / 0.005 kg → 193.8 m

Solution volume = 100 g / 1.84 g/mL → 54.3 mL

Solution volume → from mL to L → 54.3 mL . 1L/ 1000mL = 0.0543 L

Molarity → 0.969 mol / 0.0543L = 17.8 M

  • CH₃COOH → 1.05 g/mL; 99 %

Mass of solvent: 100 g - 99 g = 1 g

Moles of solvent → 1 g . 1 mol/ 18g = 0.055  moles

Mass of solute → 99 g; Moles of solute → 99 g . 1 mol/ 60 g = 1.65 moles

Total moles = 0.055 mol + 1.65 mol = 1.705 moles

Mole fraction = 1.65 / 1.705 = 0.96

Mass of solvent → from g to kg → 1 g . 1kg/1000 g = 0.001 kg

Molality → 1.65 mol / 0.001 kg → 1650 m

Solution volume = 100 g / 1.05 g/mL → 95.2mL

Solution volume → from mL to L → 95.2 mL . 1L/ 1000mL = 0.0952L

Molarity → 1.65 mol / 0.0952 L = 17.3 M

  • NH₃ → 0.90 g/mL ; 28 %

Mass of solvent: 100 g - 28 g = 72 g

Moles of solvent → 72 g . 1 mol/ 18g = 4  moles

Mass of solute → 28 g; Moles of solute → 28 g . 1mol/ 17 g = 1.64 moles

Total moles = 4 mol + 1.64 mol = 5.64 moles

Mole fraction = 1.64 / 5.64 = 0.29

Mass of solvent → from g to kg → 72 g . 1kg/1000g = 0.072kg

Molality → 1.64 mol / 0.072 kg → 22.7 m

Solution volume = 100 g / 0.90 g/mL → 111.1mL

Solution volume → from mL to L → 111.1 mL . 1L/ 1000mL = 0.111L

Molarity → 1.64 mol / 0.111L = 14.7 M

mickymike92

The molarity, molality and mole fraction of the reagents are as follows:

Hydrochloric acid:

  • Molarity = 12.4 mol/L
  • Molality = 16.77 mol/kg
  • Mole fraction = 0.23

For Nitric acid:

  • Molarity = 15.77 mol/L
  • Molality = 37 mol/Kg
  • Mole fraction = 0.40

For Sulfuric acid:

  • Molarity = 17.83 mol/L
  • Molality = 194 mol/kg
  • Mole fraction = 0.78

For Acetic acid:

  • Molarity = 17.3 mol/L
  • Molality = 1650  mol/kg
  • Mole fraction = 0.97

For Ammonia:

  • Molarity = 14.8 mol/L
  • Molality = 22.77 mol/kg
  • Mole fraction = 0.29

The formula to be used in the calculations are as follows:

  • Molarity = mass percent of solute * specific gravity / molar mass * 1000 mL/1L
  • Molality = moles of solute/mass of solvent
  • Mass of solvent = 100 - mass percent of solute
  • Mole fraction = moles of solute/ total moles of solution
  • Number of moles = mass/molar mass

For Hydrochloric acid:

density or specific gravity = 1.19

mass percent of solute = 38%

molar mass of HCl = 36.5 g/mol

moles of solute in 100 g solution = 38/36.5 = 1.04 moles

mass of solvent = 100g - 38g = 62 g = 0.062 kg

molar mass of water = 18.0 g/mol

moles of solvent in 100 g solution = 62/18 = 3.44 moles

Total moles of solution = 3.44 + 1.04 = 4.48 moles

Molarity = 0.38 * 1.19/36.5 * 1000L/1L

Molarity = 12.4 mol/L

Molality = 1.04 mol/0.062 kg

Molality = 16.77 mol/kg

Mole fraction = 1.04/4.48

Mole fraction = 0.23

For Nitric acid:

density or specific gravity = 1.42

mass percent of solute = 70%

molar mass of HNO₃ = 63.0 g/mol

moles of solute in 100 g solution = 70/63.0 = 1.11 moles

mass of solvent = 100g - 38g = 30 g = 0.030 kg

molar mass of water = 18.0 g/mol

moles of solvent in 100 g solution = 30/18 = 1.67 moles

Total moles of solution = 1.11 + 1.67 = 2.78 moles

Molarity = 0.70 * 1.42/63.0 * 1000L/1L

Molarity = 15.77 mol/L

Molality = 1.11 mol/0.030 kg

Molality = 37 mol/Kg

Mole fraction = 1.11/2.78

Mole fraction = 0.40

For Sulfuric acid:

density or specific gravity = 1.84

mass percent of solute = 95%

molar mass of H₂SO₄ = 98.0 g/mol

moles of solute in 100 g solution = 95/98.0 = 0.97 moles

mass of solvent = 100g - 95g = 5.00 g = 0.005kg

molar mass of water = 18.0 g/mol

moles of solvent in 100 g solution = 5/18 = 0.27 moles

Total moles of solution = 0.97 + 0.27 = 1.24 moles

Molarity = 0.95 * 1.84/98.0 * 1000L/1L

Molarity = 17.83 mol/L

Molality = 0.97 mol/0.005 kg

Molality = 194 mol/kg

Mole fraction = 0.97/1.24

Mole fraction = 0.78

For Acetic acid:

density or specific gravity = 1.05

mass percent of solute = 99%

molar mass of CH₃COOH = 60.0 g/mol

moles of solute in 100 g solution = 99/60 = 1.65 moles

mass of solvent = 100g - 99.0g = 1.00 g = 0.001 kg

molar mass of water = 18.0 g/mol

moles of solvent in 100 g solution = 1.00/18 = 0.055 moles

Total moles of solution = 1.65 + 0.055 = 1.705 moles

Molarity = 0.99 * 1.05/60 * 1000L/1L

Molarity = 17.3 mol/L

Molality = 1.65 mol/0.001 kg

Molality = 1650  mol/kg

Mole fraction = 1.65/1.705

Mole fraction = 0.97

For Ammonia:

density or specific gravity = 0.90

mass percent of solute = 28%

molar mass of NH₃ = 17 g/mol

moles of solute in 100 g solution = 28/17 = 1.64 moles

mass of solvent = 100g - 28g = 72 g = 0.072 kg

molar mass of water = 18.0 g/mol

moles of solvent in 100 g solution = 72/18 = 4.00 moles

Total moles of solution = 1.64 + 4.00 = 5.64 moles

Molarity = 0.28 * 0.90/17.0 * 1000L/1L

Molarity = 14.8 mol/L

Molality = 1.64 mol/0.072 kg

Molality = 22.77 mol/kg

Mole fraction = 1.64/5.64

Mole fraction = 0.29

Learn more about molarity, molality, and mole fraction at: https://brainly.com/question/10861444

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