Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 42

Write the equations relating boiling-point elevation and freezing-point depression to the concentration of the solution. Define all the terms, and give their units.

Short Answer

Expert verified
The equations for boiling-point elevation and freezing-point depression are \[ \Delta T_b = K_b \times m \] and \[ \Delta T_f = K_f \times m \] respectively. Here, \[ \Delta T_b \] and \[ \Delta T_f \] are the changes in boiling and freezing points (°C); \( K_b \) and \( K_f \) are the molal boiling-point elevation and freezing-point depression constants (°C/m); and 'm' is the molality of the solution (mol/kg).

Step by step solution

01

Definition of Terms

Boiling point elevation refers to the phenomenon whereby adding a solute to a pure solvent increases the solution’s boiling point. Freezing point depression refers to the opposite case where the solution’s freezing point is lower than that of the pure solvent. Both these phenomena are dependent on the concentration of the solution, represented by 'm', which is the molality of the solution (moles of solute/kg of solvent). The molal boiling-point elevation constant and freezing-point depression constant is denoted by 'Kb' and 'Kf' respectively. Their units are degrees Celsius per molal (°C/m).
02

Formulation of Boiling-Point Elevation Equation

The equation that relates boiling point elevation to the concentration of a solution is given by: \[ \Delta T_b = K_b \times m \] where, \[ \Delta T_b \] is the change in boiling point (°C), \( K_b \) is the molal boiling-point elevation constant (°C/m), and 'm' is the molality of the solution (mol/kg).
03

Formulation of Freezing-Point Depression Equation

The equation that relates freezing point depression to the concentration of a solution is given by: \[ \Delta T_f = K_f \times m \] where, \[ \Delta T_f \] is the change in freezing point (°C), \( K_f \) is the molal freezing-point depression constant (°C/m), and 'm' is the molality of the solution (mol/kg).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A solution of \(1.00 \mathrm{~g}\) of anhydrous aluminum chloride, \(\mathrm{AlCl}_{3}\), in \(50.0 \mathrm{~g}\) of water freezes at \(-1.11^{\circ} \mathrm{C}\). Does the molar mass determined from this freezing point agree with that calculated from the formula? Why?

Concentrated hydrochloric acid is usually available at a concentration of 37.7 percent by mass. What is its molar concentration? (The density of the solution is \(1.19 \mathrm{~g} / \mathrm{mL} . ?\)

Calculate the molality of each of the following aqueous solutions: (a) \(2.50 M \mathrm{NaCl}\) solution (density of solution \(=1.08 \mathrm{~g} / \mathrm{mL}),\) (b) 48.2 percent by

How does each of the following affect the solubility of an ionic compound? (a) lattice energy, (b) solvent (polar versus nonpolar), (c) enthalpies of hydration of cation and anion

Solutions A and B have osmotic pressures of 2.4 atm and 4.6 atm, respectively, at a certain temperature. What is the osmotic pressure of a solution prepared by mixing equal volumes of \(\mathrm{A}\) and \(\mathrm{B}\) at the same temperature?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free