Colligative Effect from Solute Mass

Introduction

A nonvolatile dissolved solute can lower a solvent's freezing point. A numerical problem often gives masses of solute and solvent rather than molality. The route is sequential: convert solute grams to moles, convert solvent grams to kilograms, calculate molality, then multiply by the solvent's freezing-point constant. Mixing up solution mass and solvent mass is the main arithmetic hazard.

Core explanation

For a dilute ideal solution of a nonelectrolyte in a specified solvent, freezing-point depression magnitude is ΔT f = K f b, where b = n solute/(kilograms of solvent). The solution freezing point is T f,solution = T f,pure − ΔT f. The depression magnitude is positive, while the final temperature shifts downward. K f has units K kg mol⁻¹ when b is mol kg⁻¹, leaving kelvin or degrees Celsius of temperature difference . A one-kelvin temperature interval equals a one-degree-Celsius interval.

If a solute's mass is m s and molar mass is M s, n s = m s/M s. If solvent mass is m w grams, solvent kilograms are m w/1000. Combining steps gives b = (m s/M s)/(m w/1000). Do not divide by total solution mass, which includes the solute and would underestimate molality. Do not use solution volume, because molality is mass-based rather than volume-based.

For water, a commonly used freezing-point depression constant is K f ≈ 1.86 K kg mol⁻¹. Suppose 9.00 g glucose, with M ≈ 180 g mol⁻¹, dissolves in 100.0 g water. Glucose moles are 0.0500 mol; solvent mass is 0.1000 kg; b = 0.500 mol kg⁻¹. Then ΔT f = (1.86)(0.500) = 0.930 K. If pure water freezes at 0.00 °C under the stated conditions, the model predicts a solution freezing point of about −0.93 °C. This is an approximate dilute-solution result, not a claim of exact behaviour for every concentration.

Colligative changes primarily reflect dissolved particle number under the ideal model. An electrolyte can dissociate into ions, requiring an effective particle factor i in ΔT f = iK fb when b is based on analytical formula units. Actual i can differ from an integer because of ion association and nonideal interactions, particularly at higher concentration. A nonelectrolyte such as glucose is usually treated as i ≈ 1 for an introductory calculation. One should not automatically assign i = 2 to any chemical formula containing two element symbols; dissociation chemistry matters.

The relation assumes solute stays dissolved and is effectively excluded from the solid solvent phase. If solute precipitates, reacts, or co-crystallises, the simple formula needs reconsideration. It also assumes a specified solvent and its K f; values for water cannot be transferred to another solvent. OpenStax Chemistry 2e's colligative-properties section provides the solvent-constant framework and its dilute-solution limitations.

Step-by-step reasoning

1. Identify solvent, solute, their masses and the solute's molar mass. 2. Convert solute mass to moles and solvent mass separately to kilograms. 3. Divide to obtain molality, then choose an appropriate effective particle factor if specified. 4. Multiply by the solvent's K f for a positive depression magnitude. 5. Subtract that magnitude from the pure-solvent freezing temperature and label the approximation.

Visual explanation

Draw a two-column table: solute 9.00 g → 0.0500 mol, solvent 100.0 g → 0.1000 kg. Merge the arrows into b = 0.500 mol kg⁻¹, then into ΔT f = K fb = 0.930 K. Finish with a thermometer showing pure solvent at 0.00 °C and solution at −0.93 °C.

Real-world analogy

Imagine distributing a fixed number of guests among a known number of hotel floors. Guests per floor depends on guest count and floor count, not on total building mass. Molality similarly counts solute units relative to solvent mass. The analogy only represents denominator choice; freezing-point change comes from thermodynamics of mixing.

Real-world example

Freezing-point measurements can help estimate a dissolved substance's particle concentration. A measured shift is compared with a known solvent constant. For a molecular solute of known mass, the result may help estimate molar mass, provided the solution is dilute, the solute remains dissolved, and no unexpected dissociation or association occurs.

Why?

Why use kilograms of solvent rather than kilograms of solution? K f is defined for molality, whose denominator counts the host solvent amount. Adding more solute changes the number of solute particles but does not redefine how much solvent was present. Using solution mass would mix numerator and denominator effects.

Common misconception

“ΔT f = −0.93 °C.” In many textbooks ΔT f denotes the positive magnitude of depression, 0.93 K; the new freezing temperature is −0.93 °C when pure water freezes at 0 °C. State the convention and avoid a double negative.

Worked example

Add 18.0 g of a nonelectrolyte with M = 90.0 g mol⁻¹ to 250.0 g water. The solute amount is 18.0/90.0 = 0.200 mol, and solvent mass is 0.2500 kg. Molality is 0.800 mol kg⁻¹. With K f(water) = 1.86 K kg mol⁻¹ and i ≈ 1, depression is ΔT f = 1.86 × 0.800 = 1.49 K. Relative to a pure-water freezing point of 0.00 °C, the predicted solution freezing point is −1.49 °C. Dividing by 268 g total solution mass would give a different and invalid molality denominator.

Quick check

1. A nonelectrolyte has 0.100 mol in 0.200 kg solvent. What is its molality? Answer: b = 0.100/0.200 = 0.500 mol kg⁻¹, using solvent rather than total solution mass.

Exam focus

Convert masses in two separate lines, show the solvent kilogram denominator, and use the constant for the specified solvent. Distinguish the positive shift magnitude from the final freezing temperature and include a particle factor only when chemically justified.

Advanced insight

Freezing-point depression results from lowering the chemical potential of liquid solvent relative to the solid solvent. The simple K fb proportionality emerges as a dilute-limit approximation. At higher concentration, solvent activity replaces an uncomplicated molality count, so deviations are expected even without a chemical reaction.

Summary

From solute mass, find moles; from solvent mass, find kilograms; divide for molality and multiply by K f, with a justified particle factor if needed. Subtract the positive depression magnitude from the pure-solvent freezing temperature. The result is a dilute-solution model prediction.

Practice questions

1. What is the molality of 0.0500 mol solute in 0.1000 kg solvent? Answer: b = 0.0500/0.1000 = 0.500 mol kg⁻¹. 2. For a nonelectrolyte in water at that molality, estimate ΔT f using K f = 1.86 K kg mol⁻¹. Answer: ΔT f = 1.86 × 0.500 = 0.930 K. 3. Why is 100 g water plus 9 g solute not treated as 0.109 kg solvent? Answer: The 9 g is solute, not solvent; solvent mass remains 0.100 kg.