Hard Molality Practice Questions

Concept Explanation

Molality is a measure of the concentration of a solute in a solution defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the total volume of the solution, molality is independent of temperature and pressure because mass does not change with thermal expansion. This makes it a critical unit in thermodynamics and for calculating colligative properties like boiling point elevation and freezing point depression. To calculate molality, you must first determine the moles of the solute using its molar mass and then divide that value by the mass of the solvent expressed in kilograms. For complex scenarios, you may need to distinguish between the mass of the solution and the mass of the solvent, often requiring the use of density to convert between units. Understanding the molarity vs molality difference is essential for students tackling advanced chemistry problems.

Solved Examples

1. Example: Converting Molarity to Molality. A 2.50 M aqueous solution of sulfuric acid (H₂SO₄) has a density of 1.15 g/mL. Calculate the molality of the solution (Molar mass of H₂SO₄ = 98.08 g/mol).

   1. Assume 1.00 L of solution. Moles of solute = 2.50 mol.

   2. Calculate mass of solution: 1000 mL × 1.15 g/mL = 1150 g.

   3. Calculate mass of solute: 2.50 mol × 98.08 g/mol = 245.2 g.

   4. Calculate mass of solvent (water): 1150 g - 245.2 g = 904.8 g = 0.9048 kg.

   5. Calculate molality: 2.50 mol / 0.9048 kg = 2.76 m.

2. Example: Molality from Percent Composition. Concentrated hydrochloric acid is 37.0% HCl by mass. What is its molality?

   1. Assume 100 g of solution. This means 37.0 g is HCl and 63.0 g is water.

   2. Convert mass of HCl to moles: 37.0 g / 36.46 g/mol = 1.015 mol.

   3. Convert mass of water to kg: 63.0 g = 0.0630 kg.

   4. Calculate molality: 1.015 mol / 0.0630 kg = 16.1 m.

3. Example: Finding Mass of Solvent. How many grams of water are needed to dissolve 50.0 g of NaCl to produce a 0.850 m solution?

   1. Calculate moles of NaCl: 50.0 g / 58.44 g/mol = 0.8556 mol.

   2. Rearrange the molality formula: mass of solvent (kg) = moles / molality.

   3. Calculate mass: 0.8556 mol / 0.850 m = 1.0066 kg.

   4. Convert to grams: 1.0066 kg × 1000 = 1006.6 g H₂O.

Practice Questions

1. A solution is prepared by dissolving 45.0 g of glucose (C₆H₁₂O₆) in 150.0 g of water. Calculate the molality of the sugar solution.

2. A 3.20 m aqueous solution of ethanol (C₂H₅OH) is required for an experiment. If you have 500.0 g of water, how many grams of ethanol must you add?

3. An aqueous solution of phosphoric acid (H₃PO₄) is 25.0% by mass and has a density of 1.15 g/mL. Determine the molality.

4. A sample of seawater contains 3.5% sodium chloride by mass. Assuming the rest is pure water, calculate the molality of NaCl in the ocean water.

5. A 1.45 M solution of nitric acid (HNO₃) has a density of 1.08 g/mL. Calculate its molality.

6. If you mix 100.0 mL of 2.00 m KCl (density = 1.10 g/mL) with 200.0 mL of water, what is the new molality of the resulting solution?

7. A solution is prepared by mixing 25.0 mL of methanol (CH₃OH, density = 0.791 g/mL) with 100.0 mL of ethanol (C₂H₅OH, density = 0.789 g/mL). Calculate the molality of methanol in ethanol.

8. What is the molality of a solution containing 12.8 g of naphthalene (C₁₀H₈) dissolved in 50.0 g of benzene (C₆H₆)?

9. A sulfuric acid solution has a molality of 4.50 m. Calculate the mass percent of H₂SO₄ in this solution.

10. If the molality of an aqueous solution of a non-electrolyte is 1.25 m, and the solution contains 250 g of water, how many grams of solute are present if the molar mass of the solute is 180.16 g/mol?

Answers & Explanations

1. Answer: 1.67 m.
   Moles of glucose = 45.0 g / 180.16 g/mol = 0.2498 mol. Mass of solvent = 150.0 g = 0.150 kg. Molality = 0.2498 mol / 0.150 kg = 1.67 m.

2. Answer: 73.7 g.
   Moles needed = 3.20 mol/kg × 0.500 kg = 1.60 mol. Mass = 1.60 mol × 46.07 g/mol = 73.71 g.

3. Answer: 3.40 m.
   In 100 g solution: 25 g H₃PO₄ and 75 g water. Moles H₃PO₄ = 25 / 98.0 g/mol = 0.255 mol. Molality = 0.255 mol / 0.075 kg = 3.40 m. (Note: Density is extraneous information here).

4. Answer: 0.62 m.
   In 100 g seawater: 3.5 g NaCl and 96.5 g water. Moles NaCl = 3.5 / 58.44 = 0.0599 mol. Molality = 0.0599 / 0.0965 kg = 0.621 m.

5. Answer: 1.47 m.
   1 L solution = 1080 g. Moles HNO₃ = 1.45 mol. Mass HNO₃ = 1.45 × 63.01 = 91.36 g. Mass solvent = 1080 - 91.36 = 988.64 g = 0.9886 kg. Molality = 1.45 / 0.9886 = 1.467 m.

6. Answer: 0.733 m.
   First, find moles in 100 mL of 2.00 m KCl. Mass of solution = 110 g. Let x = mass of solute. Moles = x/74.55. Molality = (x/74.55) / [(110-x)/1000] = 2.00. Solving for x ≈ 14.28 g KCl. Moles = 0.1916. Original solvent = 110 - 14.28 = 95.72 g. New solvent = 95.72 + 200 = 295.72 g = 0.2957 kg. New molality = 0.1916 / 0.2957 = 0.648 m.

7. Answer: 7.84 m.
   Mass methanol = 25.0 × 0.791 = 19.775 g. Moles methanol = 19.775 / 32.04 = 0.6172 mol. Mass ethanol (solvent) = 100.0 × 0.789 = 78.9 g = 0.0789 kg. Molality = 0.6172 / 0.0789 = 7.82 m.

8. Answer: 2.00 m.
   Moles naphthalene = 12.8 / 128.17 = 0.09987 mol. Mass benzene = 0.050 kg. Molality = 0.09987 / 0.050 = 1.997 m.

9. Answer: 30.6%.
   4.50 mol H₂SO₄ in 1 kg water. Mass H₂SO₄ = 4.50 × 98.08 = 441.36 g. Total mass = 1000 + 441.36 = 1441.36 g. Mass % = (441.36 / 1441.36) × 100 = 30.6%.

10. Answer: 56.3 g.
    Moles = 1.25 mol/kg × 0.250 kg = 0.3125 mol. Mass = 0.3125 mol × 180.16 g/mol = 56.3 g.

Quick Quiz

Frequently Asked Questions

What is the difference between molarity and molality?

Molarity is defined as moles of solute per liter of solution, making it temperature-dependent as liquids expand or contract. Molality is moles of solute per kilogram of solvent, which remains constant regardless of temperature or pressure changes.

Why is molality used in freezing point depression?

Molality is used because colligative properties depend on the ratio of solute particles to solvent molecules. Since mass does not change with temperature, using molality ensures that the concentration value remains accurate throughout the freezing or boiling process.

Can molality be higher than molarity?

Yes, molality is typically higher than molarity in aqueous solutions because the mass of 1 liter of solution is usually greater than the mass of the solvent alone, but the volume of the solution is greater than the volume of the solvent. For very dilute aqueous solutions, the values are nearly identical.

How do you calculate molality from mass percent?

To calculate molality from mass percent, assume a 100g sample to find the mass of solute and solvent. Convert the solute mass to moles and the solvent mass to kilograms, then divide the moles by the kilograms.

Does the type of solvent affect molality?

The identity of the solvent affects the calculation only through its mass; however, in practice, different solvents have different molar masses and densities which must be accounted for when preparing solutions by volume.

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