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Lesson: Cryoscopy and Ebullioscopy

In this lesson, we will learn how to calculate the molar mass of a solute based on its effect on the melting or boiling point of the solvent.

Worksheet • 8 Questions

Q1:

A 12.0 g sample of a non-electrolyte is dissolved in 80.0 g of water. The solution freezes at − 1 . 9 4 ∘ C . What is the molar mass of this solute? The freezing point depression constant of water is 1 . 8 6 / ∘ C m .

Q2:

Why are soaps useful for cleaning?

  • ASoaps are amphiphilic molecules that help to emulsify hydrophobic substances, allowing them to be removed by water.
  • BSoaps lower the viscosity of water. This allows it to penetrate hydrophobic substances more easily, breaking them into smaller particles that can be mechanically removed.
  • CSoaps are very reactive and decompose grease into smaller particles that can be dissolved by water.
  • DSoap molecules function as “molecular sandpaper”: the hydrophobic tails of the molecules are inserted between molecules of an insoluble substance, mechanically separating them into smaller particles that can be dissolved by water.
  • ESoaps are absorbed into grease and other hydrophobic substances. This causes the materials to swell and become less sticky, allowing them to removed by water.

Q3:

Lysozyme is an enzyme that cleaves cell walls. A 0.100 L sample of a solution of lysozyme containing 0.0750 g of the enzyme exhibits an osmotic pressure of 1 . 3 2 × 1 0 − 3 atm at 2 5 ∘ C . What is the molar mass of lysozyme?

  • A 1 . 3 9 × 1 0 4 g/mol
  • B 6 . 0 8 × 1 0 4 g/mol
  • C 1 . 3 2 × 1 0 4 g/mol
  • D 4 . 5 0 × 1 0 4 g/mol
  • E 3 . 4 5 × 1 0 3 g/mol

Q4:

A sample of an organic compound (a non-electrolyte) weighing 1.30 g lowered the freezing point of 10.0 g of benzene ( 𝐾 = 5 . 1 2 / ) f ∘ C m by 3 . 6 6 ∘ C . What is the molar mass of the compound?

Q5:

When 5.00 g of a non-ionic compound is dissolved in 25.00 g of carbon tetrachloride (boiling point 7 6 . 8 ∘ C , 𝐾 = 5 . 0 2 ⋅ / ) b K k g m o l , the boiling point of the solution at 1 atm is 8 1 . 5 ∘ C . Calculate the molar mass of the non-ionic compound.

Q6:

A sample of sulfur weighing 0.201 g was dissolved in 17.8 g of carbon disulfide, C S 2   ( 𝐾 = 2 . 4 3 / ) b ∘ C m . If the boiling point of the carbon disulfide was elevated by 0 . 1 0 7 ∘ C , what is the formula of a sulfur molecule in this solution?

  • A S 8
  • B S 5
  • C S
  • D S 4
  • E S 2

Q7:

A 2.43 mol/kg aqueous solution of glycerin ( C H O ) 3 8 3 was prepared by mixing glycerin with 0.500 kg of water.

What mass of glycerin was required?

What is the freezing point of this solution? The freezing point depression constant of water is 1.86 kg⋅K/mol.

Q8:

When 6.29 g of a non-volatile solute is dissolved in 500 g of water, the freezing point of the resultant aqueous solution is 0 . 6 4 6 ∘ C lower than that of pure water. The cryoscopic constant for water is 𝐾 = 1 . 8 5 6 ⋅ / 𝑓 K k g m o l and the solute does not dissociate. Estimate the molar mass of the solute.
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