/*! 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 7 Using the phase diagram for wate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 7

Using the phase diagram for water and Raoult's law, explain why salt is spread on the roads in winter (even when it is below freezing).

Short Answer

Expert verified
When salt is spread on roads during winter, it lowers the freezing point of water due to the freezing point depression, as explained by Raoult's law. This law states that the vapor pressure of a solvent in a solution decreases with the increase in solute concentration. In this case, adding salt (solute) to water (solvent) results in a solution with a lower freezing point than pure water. Consequently, the water on the roads remains liquid at lower temperatures, preventing ice formation and making the roads less slippery and safer for driving.

Step by step solution

01

Understanding the Phase Diagram for Water

A phase diagram is a graphical representation of the different phases (solid, liquid, and gas) of a substance under varying conditions, such as temperature and pressure. In this exercise, we are focusing on the phase diagram for water. The most crucial point in the diagram is the freezing point (0掳C, 32掳F), which is when water turns to ice. As the temperature decreases, water begins to freeze, and ice forms. This is what occurs on roads during winter when the temperature drops, causing ice and slippery conditions.
02

Understanding Raoult's Law

Raoult's law states that the vapor pressure of a solvent in a solution decreases as the concentration of solute increases. In simpler terms, adding solute (like salt) to a solvent (like water) will lower the vapor pressure of the solvent. This means that a solution of water and salt will have a lower vapor pressure compared to pure water. This decrease in vapor pressure affects the freezing point of the solution, which is known as "freezing point depression."
03

Applying Raoult's Law to the Phase Diagram

When salt is added to water, the freezing point temperature of the water decreases due to freezing point depression. In other words, the solution of salt and water will freeze at a lower temperature than pure water. This effect can be observed in the phase diagram of water since it shifts the freezing point curve to lower temperatures. This means that the water on the road will remain liquid, rather than turning into ice, at a lower temperature than if no salt was added.
04

Explaining the Phenomenon

The use of salt in winter road treatment takes advantage of the freezing point depression caused by the presence of solute (salt) in the solvent (water). When salt is spread on the roads, it mixes with the existing water, forming a solution with a lower freezing point than pure water. The freezing point depression allows the water to remain in a liquid state, even in temperatures below the freezing point of pure water. This liquid state prevents ice formation on the road, making it less slippery and safer for vehicles to drive.

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 91影视!

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 volume of a \(0.580 M\) solution of \(\mathrm{CaCl}_{2}\) contains \(1.28 \mathrm{~g}\) solute?

An aqueous solution of \(10.00 \mathrm{~g}\) of catalase, an enzyme found in the liver, has a volume of \(1.00 \mathrm{~L}\) at \(27^{\circ} \mathrm{C}\). The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be \(0.745\) torr. Calculate the molar mass of catalase.

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When \(1.00 \mathrm{~g}\) reserpine is dissolved in \(25.0 \mathrm{~g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\) for camphor is \(40 .{ }^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) ). Calculate the molality of the solution and the molar mass of reserpine.

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \mathrm{X}\) \(10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration \((\mathrm{mol} / \mathrm{L})\).

What volume of \(0.25 M \mathrm{HCl}\) solution must be diluted to prepare \(1.00 \mathrm{~L}\) of \(0.040 \mathrm{M} \mathrm{HCl} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

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.