BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Solutions

Chapter
Section
BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation.)

a. 0.050 m MgCl2

b. 0.050 m FeCl3

Interpretation Introduction

Interpretation: The freezing point and the boiling point of the given solutions are should be calculated.

Concept introduction:

  • Elevation of boiling point:

    The boiling point of the solution is increases when the solute is dissolved in the solvent is called Elevation of boiling point. it is one of the colligative Properties thus,

                                            ΔT=iKbmsolute......(1)ΔT is boiling-point elevationKbismolal boiling-point elevation constantmis molality of the soluteiisthevan't Hoff factor

  • Depression in freezing point:

    The freezing point the solution is decreases when the solute is dissolved in the solvent is called depression in freezing point. it is one of the colligative Properties thus,

                                 ΔT=iKfmsolute......(2)ΔT is boiling-point elevationKfismolal freezing-point depression constantmis molality of the soluteiisthevan't Hoff factor

Explanation

Record the given data,

           Molality of MgCl2= 0.050m

           Molality of FeCl3=0.050m

To calculate the boiling point of 0.050mMgCl2 solution.

Molal boiling-point elevation constant of water is 0.51 °C/molal .

                    i=moleofionsmoleofsolutei=3.0/1ΔT=3.0×0.51 °C/molal×0.050molal=0.077°C=100°C+0.077°CTb=100.077°C

  • The given values are plugging in to equation 1 to give elevation of boiling point this was added to boiling point of (solvent) water (100°C) to give the boiling point of 0.050mMgCl2 solution.
  • The boiling point of 0.050mMgCl2 solution is 100.077°C

To calculate the freezing point of 0.050mMgCl2 solution

Molal freezing-point depression constantof water is 1.86°C/molal

                                 i=moleofionsmoleofsolutei=3.0/1ΔT=3.0×1.86 °C/molal×0.050molal=0.28°CTf=0.00°C-0.28°C=-28.0°C

  • The given values are plugging in to equation 1 to give depression in freezing point this was subtract from freezing point of (solvent) water (100°C) to give the boiling point of 0.050mMgCl2 solution.
  • The freezing point of 0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 11 Solutions

Show all chapter solutions add
Ch-11 P-1ALQCh-11 P-2ALQCh-11 P-3ALQCh-11 P-4ALQCh-11 P-5ALQCh-11 P-6ALQCh-11 P-7ALQCh-11 P-8ALQCh-11 P-9ALQCh-11 P-10ALQCh-11 P-11SRCh-11 P-12SRCh-11 P-13SRCh-11 P-14SRCh-11 P-15SRCh-11 P-16SRCh-11 P-17QCh-11 P-18QCh-11 P-19QCh-11 P-20QCh-11 P-21QCh-11 P-22QCh-11 P-23QCh-11 P-24QCh-11 P-25QCh-11 P-26QCh-11 P-27QCh-11 P-28QCh-11 P-29ECh-11 P-30ECh-11 P-31ECh-11 P-32ECh-11 P-33ECh-11 P-34ECh-11 P-35ECh-11 P-36ECh-11 P-37ECh-11 P-38ECh-11 P-39ECh-11 P-40ECh-11 P-41ECh-11 P-42ECh-11 P-43ECh-11 P-44ECh-11 P-45ECh-11 P-46ECh-11 P-47ECh-11 P-48ECh-11 P-49ECh-11 P-50ECh-11 P-51ECh-11 P-52ECh-11 P-53ECh-11 P-54ECh-11 P-55ECh-11 P-56ECh-11 P-57ECh-11 P-58ECh-11 P-59ECh-11 P-60ECh-11 P-61ECh-11 P-62ECh-11 P-63ECh-11 P-64ECh-11 P-65ECh-11 P-66ECh-11 P-67ECh-11 P-68ECh-11 P-69ECh-11 P-70ECh-11 P-71ECh-11 P-72ECh-11 P-73ECh-11 P-74ECh-11 P-75ECh-11 P-76ECh-11 P-77ECh-11 P-78ECh-11 P-79ECh-11 P-80ECh-11 P-81ECh-11 P-82ECh-11 P-83ECh-11 P-84ECh-11 P-85ECh-11 P-86ECh-11 P-87ECh-11 P-88ECh-11 P-89ECh-11 P-90ECh-11 P-91AECh-11 P-92AECh-11 P-94AECh-11 P-95AECh-11 P-96AECh-11 P-97AECh-11 P-98AECh-11 P-99AECh-11 P-100AECh-11 P-101AECh-11 P-102AECh-11 P-103AECh-11 P-104AECh-11 P-105AECh-11 P-106AECh-11 P-107AECh-11 P-108AECh-11 P-109AECh-11 P-110CWPCh-11 P-111CWPCh-11 P-112CWPCh-11 P-113CWPCh-11 P-114CWPCh-11 P-115CWPCh-11 P-116CWPCh-11 P-117CWPCh-11 P-118CPCh-11 P-119CPCh-11 P-120CPCh-11 P-121CPCh-11 P-122CPCh-11 P-123CPCh-11 P-124CPCh-11 P-125CPCh-11 P-126CPCh-11 P-127CPCh-11 P-128CPCh-11 P-129CPCh-11 P-130IPCh-11 P-131IPCh-11 P-132IP

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

A person who exercises moderately for longer than 20 minutes begins to a. use less glucose and more fat for fue...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Why are some quantities called fundamental?

An Introduction to Physical Science

Why does the orbital period of a binary star depend on its mass?

Horizons: Exploring the Universe (MindTap Course List)

Chloroplasts of green algae evolved from _____.

Biology: The Unity and Diversity of Life (MindTap Course List)

A GeigerMueller tube is a radiation detector that consists of a closed, hollow, metal cylinder (the cathode) of...

Physics for Scientists and Engineers, Technology Update (No access codes included)