Unknown Solute from Multiple Properties

Introduction

Suppose a laboratory receives a clear, colourless solution with only a label giving the mass of solute dissolved. No single measurement will reveal what the solute is, but several simple ones together can. A density reading, a freezing-point depression and an osmotic-pressure value each describe the same solution from a different angle. This page shows how to link them into one consistent calculation, how to use each result as a check on the others, and why an extra piece of information is sometimes needed to finish the identification.

Core explanation

What colligative data really count. Freezing-point depression and osmotic pressure respond to the number of dissolved particles. A measurement therefore gives the amount of particles, not directly the amount of formula units. For a non-electrolyte such as a sugar, the two are the same (i = 1). For a salt such as NaCl, each formula unit gives about two particles (i ≈ 2).

Freezing-point depression works on a mass basis. The relationship is ΔTf = i·Kf·b, where b is molality in mol kg⁻¹ of solvent. Because molality uses the mass of solvent, you need only the masses weighed out; no volume is involved. Rearranging gives the particle molality, and multiplying by the solvent mass in kilograms gives the amount of particles.

Osmotic pressure works on a volume basis. The relationship is Π = i·c·R·T, where c must be in mol m⁻³ when Π is in pascals and R = 8.314 J K⁻¹ mol⁻¹. Here the concentration is per unit volume of solution, so you must know the solution volume.

Density is the bridge. A solution made from known masses of solute and solvent has a known total mass. Dividing by the measured density gives its volume. That volume converts molality into molarity and lets the two colligative measurements be compared directly.

Apparent molar mass. Dividing the mass of solute by the amount of particles gives an apparent molar mass. If i = 1 this equals the true molar mass. If the solute dissociates, the apparent value is too small by a factor of i. Colligative data alone cannot tell the two cases apart, so a further clue — conductivity, elemental composition or chemical behaviour — decides between them.

Which method suits which solute? For small molecules the freezing shift is large enough to measure well. For large molecules such as proteins, the amount of particles is tiny and the freezing shift becomes too small to read, but osmotic pressure is still measurable, so it is the preferred method for macromolecules.

Formulae

ΔTf = i·Kf·b, with b = n(particles) ÷ m(solvent in kg)

Π = i·c·R·T, with c in mol m⁻³, Π in Pa, T in K

V(solution) = [m(solute) + m(solvent)] ÷ ρ(solution)

Apparent M = m(solute) ÷ n(particles)

Step-by-step reasoning

1. From ΔTf and Kf, find the particle molality, then the amount of particles. 2. From the masses and the density, find the solution volume in m³. 3. From Π, find the particle concentration c = Π ÷ RT, then the amount of particles in that volume. 4. Check that steps 1 and 3 agree within experimental uncertainty. 5. Divide the solute mass by the amount of particles to obtain the apparent molar mass. 6. Use conductivity or composition data to decide the value of i and the formula.

Visual explanation

Draw three boxes in a row labelled "freezing point", "density" and "osmotic pressure". Arrows from the first and last boxes both point to a central circle labelled "amount of particles". The density box sits on the arrow from osmotic pressure, showing that it is needed to convert a volume-based concentration into an amount. A final arrow from the circle to "molar mass" is marked "÷ i?".

Real-world analogy

Counting people in a stadium by two methods — ticket scans at the gates and a head count from aerial photographs — should give the same number. If the two counts agree, you trust them. Neither count, however, tells you whether the people came as individuals or in pairs; you need a separate clue, such as the type of ticket sold.

Real-world example

Biochemists use membrane osmometry to estimate the molar masses of proteins and synthetic polymers. A solution containing only a few grams of protein per litre produces an osmotic pressure of a few hundred pascals — easily measured — while its freezing-point depression would be well below a thousandth of a kelvin.

Why?

Why do both measurements count particles rather than mass? Freezing and osmosis are governed by how much the solvent is diluted: dissolved particles lower the chemical potential of the solvent in proportion to their mole fraction. One heavy molecule dilutes the solvent exactly as much as one light ion does, so identity drops out and only the number of particles matters.

Common misconception

"A colligative measurement gives the molar mass of the solute directly." It gives the amount of particles. If the solute dissociates or associates, the result must be corrected by i before it represents the formula-unit molar mass.

Worked example

Question: A solution of 9.00 g of an unknown solid in 100.0 g of water freezes at −0.93 °C and has a density of 1.03 g cm⁻³. At 298 K its osmotic pressure is 1.17 × 10⁶ Pa. The solid does not conduct electricity in solution, and analysis shows the empirical formula CH₂O. Identify the solute. (Kf = 1.86 K kg mol⁻¹.)

Reasoning: Freezing data: b = 0.93 ÷ 1.86 = 0.500 mol kg⁻¹, so n(particles) = 0.500 × 0.1000 = 0.0500 mol.

Density: mass of solution = 109.0 g; V = 109.0 ÷ 1.03 = 105.8 cm³ = 1.058 × 10⁻⁴ m³.

Osmotic data: c = Π ÷ RT = 1.17 × 10⁶ ÷ (8.314 × 298) = 472 mol m⁻³; n(particles) = 472 × 1.058 × 10⁻⁴ = 0.0499 mol. The two methods agree.

Apparent M = 9.00 ÷ 0.0500 = 180 g mol⁻¹. The solution does not conduct, so i = 1 and M = 180 g mol⁻¹. The empirical formula mass of CH₂O is 30, and 180 ÷ 30 = 6.

Answer: The molecular formula is C₆H₁₂O₆, a hexose sugar such as glucose.

Quick check

1. Why does the osmotic-pressure route need the solution density, while the freezing-point route does not? Answer: Osmotic pressure depends on concentration per volume of solution, and density converts the known mass of solution into that volume; freezing depression uses molality, which needs only masses.

Exam focus

Keep units consistent: Π in Pa with c in mol m⁻³, or Π in kPa with c in mol dm⁻³. State clearly whether you have found an amount of particles or of formula units. Examiners reward an explicit consistency check between two independent measurements.

Advanced insight

Real solutions deviate from ideal behaviour. For electrolytes, ion pairing makes the measured i smaller than the whole-number value, so 0.1 mol kg⁻¹ NaCl behaves as if i ≈ 1.9. For polymers, Π ÷ c is plotted against c and extrapolated to zero concentration, because interactions between large molecules make the simple equation inaccurate even in fairly dilute solutions.

Summary

Freezing-point depression gives the amount of dissolved particles from mass data, while osmotic pressure gives it from a volume-based concentration. Solution density converts between the two bases, so the results can be cross-checked. Dividing the solute mass by the particle amount gives an apparent molar mass, which equals the true molar mass only when i = 1. Conductivity or composition data settle the value of i and complete the identification.

Practice questions

1. 2.00 g of a non-electrolyte dissolved in 50.0 g of water lowers the freezing point by 0.40 K. Calculate its molar mass. (Kf = 1.86 K kg mol⁻¹) Answer: b = 0.40 ÷ 1.86 = 0.215 mol kg⁻¹; n = 0.215 × 0.0500 = 0.0108 mol; M = 2.00 ÷ 0.0108 ≈ 186 g mol⁻¹. 2. 1.00 g of a protein in 100 cm³ of solution gives an osmotic pressure of 250 Pa at 298 K. Estimate its molar mass. Answer: c = 250 ÷ (8.314 × 298) = 0.101 mol m⁻³; n = 0.101 × 1.00 × 10⁻⁴ = 1.01 × 10⁻⁵ mol; M ≈ 1.00 ÷ 1.01 × 10⁻⁵ ≈ 9.9 × 10⁴ g mol⁻¹. 3. An unknown salt gives an apparent molar mass of 29 g mol⁻¹ from freezing data, and its solution conducts strongly. Suggest a formula-unit molar mass and a possible identity. Answer: A 1:1 salt gives i ≈ 2, so M ≈ 58 g mol⁻¹, consistent with sodium chloride (58.4 g mol⁻¹). 4. Why is osmotic pressure preferred to freezing-point depression for the protein in question 2? Answer: The amount of particles is so small that the freezing shift would be only about 0.0002 K, far too small to measure accurately, whereas 250 Pa is easily measured.