/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 What is the freezing point of \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the freezing point of \(0.0075 \mathrm{~m}\) aqueous calcium chloride, \(\mathrm{CaCl}_{2}\) ? Use the formula of the salt to obtain \(i\).

Short Answer

Expert verified
The freezing point is approximately \(-0.042\, ^{\circ}\text{C}\).

Step by step solution

01

Understanding the Problem

We need to find the freezing point of a 0.0075 m (molality) aqueous solution of calcium chloride, \(\mathrm{CaCl}_{2}\). We'll use the formula for freezing point depression, \(\Delta T_f = i \cdot K_f \cdot m\), where \(i\) is the van 't Hoff factor, \(K_f\) is the freezing point depression constant for water, and \(m\) is the molality of the solution.
02

Determine the van 't Hoff Factor \(i\)

Calcium chloride dissociates into three ions in solution: one \(\mathrm{Ca}^{2+}\) ion and two \(\mathrm{Cl}^-\) ions. Therefore, the van 't Hoff factor \(i\) is 3.
03

Identify the Freezing Point Depression Constant \(K_f\)

For water, the freezing point depression constant \(K_f\) is approximately \(1.86 \, ^{\circ}\text{C/m}\).
04

Apply the Freezing Point Depression Formula

Substitute the values into the formula: \(\Delta T_f = i \cdot K_f \cdot m = 3 \cdot 1.86 \, ^{\circ}\text{C/m} \cdot 0.0075 \, \mathrm{m}\).
05

Calculate \(\Delta T_f\)

Calculate \(\Delta T_f\): \(\Delta T_f = 3 \times 1.86 \times 0.0075 = 0.04185 \, ^{\circ}\text{C}\).
06

Determine the Freezing Point of the Solution

The freezing point of pure water is \(0 \, ^{\circ}\text{C}\). The freezing point of the solution is \(0 \, ^{\circ}\text{C} - \Delta T_f = 0 - 0.04185 \, ^{\circ}\text{C} = -0.04185 \, ^{\circ}\text{C}\).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

van 't Hoff factor
The van 't Hoff factor, often represented by the symbol \(i\), is crucial in understanding how solutes affect the colligative properties of solutions. Essentially, it is the number of particles into which a compound dissociates in solution. For example, calcium chloride (\(\text{CaCl}_2\)) dissociates into three ions: one calcium ion (\(\text{Ca}^{2+}\)) and two chloride ions (\(\text{Cl}^-\)).

Therefore, for calcium chloride in solution, \(i = 3\). This implies that each formula unit of \(\text{CaCl}_2\) results in three ions.

The van 't Hoff factor helps calculate the degree of change in boiling point, freezing point, and osmotic pressure of a solution. Understanding this factor is vital when predicting how certain applications, like anti-freeze or laboratory scenarios, will react when mixed with water or another solvent.
aqueous solution
An aqueous solution is a mixture in which water serves as the solvent. In our context, calcium chloride is dissolved in water, resulting in an aqueous solution.

Water, being a polar solvent, is very effective at dissolving ionic compounds like calcium chloride (\(\text{CaCl}_2\)). Upon dissolution, water molecules surround and separate the ions present in the solid compound, which leads to the complete dissociation of \(\text{CaCl}_2\) into its constituent ions \((\text{Ca}^{2+}\) and \(\text{Cl}^-)\).

Aqueous solutions are integral in studying reactions, especially in chemistry, as they can provide insights into molecular interactions, solution concentration, and the effects of varying substance amounts.
molality
Molality (\(m\)) is a measure of the concentration of a solute in a solution, expressed in moles of solute per kilogram of solvent. Unlike molarity, molality is not temperature-dependent since it is based on the mass of the solvent rather than its volume. This makes it particularly useful when dealing with temperature variations in experimental settings.

For example, when we talk about the 0.0075 \(m\) aqueous solution of calcium chloride, we mean there are 0.0075 moles of \(\text{CaCl}_2\) dissolved per kilogram of water.

This unit of concentration is essential for calculating colligative properties such as freezing point depression, because it takes into account changes due to temperature and is a more stable representation than molarity.
calcium chloride
Calcium chloride, chemically represented as \(\text{CaCl}_2\), is a white crystalline salt known for its high solubility in water. When calcium chloride is dissolved in an aqueous medium, it dissociates into one calcium ion \((\text{Ca}^{2+})\) and two chloride ions \((\text{Cl}^-)\).

This property makes \(\text{CaCl}_2\) very effective in applications such as de-icing roads, where it is used to lower the freezing point of water. Calcium chloride's ability to attract moisture from the environment is leveraged in drying agents and dust control measures.

Due to its ionic nature, \(\text{CaCl}_2\) is also heavily applied in medical treatments as an electrolyte, in food preservation, and even in the construction industry to accelerate the setting time of concrete.

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

What mass of solution containing \(5.00 \%\) potassium iodide, KI, by mass contains \(258 \mathrm{mg}\) KI?

Ten grams of the hypothetical ionic compounds \(\mathrm{XZ}\) and YZ are each placed in a separate \(2.0-\mathrm{L}\) beaker of water. \(\mathrm{XZ}\) completely dissolves, whereas YZ is insoluble. The energy of hydration of the \(\mathrm{Y}^{+}\) ion is greater than the \(\mathrm{X}^{+}\) ion. Explain this difference in solubility.

A \(0.0140-\mathrm{g}\) sample of an ionic compound with the formula \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}_{3}\) was dissolved in water to give \(25.0 \mathrm{~mL}\) of solution at \(25^{\circ} \mathrm{C}\). The osmotic pressure was determined to be \(119 \mathrm{mmHg}\). How many ions are obtained from each formula unit when the compound is dissolved in water?

The solubility of carbon dioxide in water is \(0.161 \mathrm{~g} \mathrm{CO}_{2}\) in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\). A soft drink is carbonated with carbon dioxide gas at \(5.50\) atm pressure. What is the solubility of carbon dioxide in water at this pressure?

What is the total vapor pressure at \(20^{\circ} \mathrm{C}\) of a liquid solution containing \(0.30\) mole fraction benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), and \(0.70\) mole fraction toluene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) ? Assume that Raoult's law holds for each component of the solution. The vapor pressure of pure benzene at \(20^{\circ} \mathrm{C}\) is \(75 \mathrm{mmHg}\); that of toluene at \(20^{\circ} \mathrm{C}\) is \(22 \mathrm{mmHg}\).

See all solutions

Recommended explanations on Chemistry Textbooks

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.