The molarity, molality and mole fraction of the reagents are as follows:
Hydrochloric acid:
Molarity = 12.4 mol/LMolality = 16.77 mol/kgMole fraction = 0.23
For Nitric acid:
Molarity = 15.77 mol/LMolality = 37 mol/KgMole fraction = 0.40
For Sulfuric acid:
Molarity = 17.83 mol/LMolality = 194 mol/kgMole fraction = 0.78
For Acetic acid:
Molarity = 17.3 mol/L Molality = 1650 mol/kgMole fraction = 0.97
For Ammonia:
Molarity = 14.8 mol/L Molality = 22.77 mol/kgMole 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/1LMolality = moles of solute/mass of solventMass of solvent = 100 - mass percent of soluteMole fraction = moles of solute/ total moles of solutionNumber 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
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