Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 17E
Interpretation Introduction

(a)

Interpretation:

A concept map is to be drawn and the liters of HF gas dissolved in the solution when 1.00 g of hydrogen fluoride dissolves in 100 mL of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as mol Mole is defined that the amount of substance that contains molecules or atoms equals to 12g of C12 molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Expert Solution
Check Mark

Answer to Problem 17E

The concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 17E , additional homework tip  1

The liters of HF gas dissolved in the solution when 1.00 g of hydrogen fluoride dissolves in 100 mL of solution is 1.12L.

Explanation of Solution

When 1.00 g of hydrogen fluoride dissolves in 100 mL of solution, mole concept map looks like as shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 17E , additional homework tip  2

Figure 1

The formula to calculate the number of moles of HF gas is given below.

Numberofmoles=MassofHFMolarmassofHF

The mass of HF gas is 1.00 g.

The molar mass of HF gas is 20.01gmol1.

Substitute the mass and molar mass of HF gas in the above equation.

Numberofmoles=1.00 g20.01gmol1=0.05mol

The volume occupied by 1mol of HF gas at STP is 22.4L.

The formula to calculate the volume occupied by 0.05mol of HF gas is shown below.

Volume= 0.05molHF×Volumeof1molHF1molHF

Substitute the volume of 1mol of HF gas in the above equation.

Volume= 0.05molHF×22.4L1molHF=1.12L

Therefore, the liters of HF gas dissolved in the solution is 1.12L.

Conclusion

The liters of HF gas is 1.12L.

Interpretation Introduction

(b)

Interpretation:

A concept map is to be drawn and the molecules of HF gas dissolved in the solution in when 1.00 g of hydrogen fluoride dissolves in 100 mL of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as mol Mole is defined that the amount of substance that contains molecules or atoms equals to 12g of C12 molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Expert Solution
Check Mark

Answer to Problem 17E

The concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 17E , additional homework tip  3

The molecules of HF gas dissolved in the solution in when 1.00 g of hydrogen fluoride dissolves in 100 mL of solution is 3.01×1022moleules.

Explanation of Solution

When 1.00 g of hydrogen fluoride dissolves in 100 mL of solution, mole concept map looks like as shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 17E , additional homework tip  4

Figure 1

The formula to calculate the number of moles of HF gas is given below.

Numberofmoles=MassofHFMolarmassofHF

The mass of HF gas is 1.00 g.

The molar mass of HF gas is 20.01gmol1.

Substitute the mass and molar mass of HF gas in the above equation.

Numberofmoles=1.00 g20.01gmol1=0.05mol

The molecules present in 1mol of HF gas at STP are 6.02×1023moleules.

The formula to calculate the molecules occupied by 0.05mol of HF gas is shown below.

Numberofmoleucles= 0.05molHF×Moleculesin1molHF1molHF

Substitute the molecules in 1mol of HF gas in the above equation.

Numberofmoleucles= 0.05molHF×6.02×1023moleules1molHF=3.01×1022moleules

Therefore, the molecules of HF gas dissolved in the solution is 3.01×1022moleules.

Conclusion

The molecules of HF gas is 3.01×1022moleules.

Interpretation Introduction

(c)

Interpretation:

A concept map is to be drawn and molar concentration of the hydrofluoric solution in the solution in when 1.00 g of hydrogen fluoride dissolves in 100 mL of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as mol Mole is defined that the amount of substance that contains molecules or atoms equals to 12g of C12 molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Expert Solution
Check Mark

Answer to Problem 17E

The concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 17E , additional homework tip  5

The molar concentration of the hydrofluoric acid solution is 0.5M.

Explanation of Solution

When 1.00 g of hydrogen fluoride dissolves in 100 mL of solution, mole concept map looks like as shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 17E , additional homework tip  6

Figure 1

The formula to calculate the number of moles of HF gas is given below.

Numberofmoles=MassofHFMolarmassofHF

The mass of HF gas is 1.00 g.

The molar mass of HF gas is 20.01gmol1.

Substitute the mass and molar mass of HF gas in the above equation.

Numberofmoles=1.00 g20.01gmol1=0.05mol

The number of moles in 100 mL of solution is 0.05mol.

The relation between L and mL is shown below.

1L=1000mL

The probable unit factors are given below.

1L1000mLand1000mL1L

The unit factor to determine L from mL is given below.

1L1000mL

Therefore, the volume in L is calculated below.

Volume=100 mL×1L1000mL=0.1L

The formula to determine molarity is shown below.

M=nV …(1)

Where

M is the molarity of a solution.

n is the number of moles of solute.

V is the volume of the solution.

Substitute the value of number of moles and volume in equation (1).

MolarconcentrationofHF=0.05mol0.1 L=0.5mol/L

The relation between M and mol/L as shown below.

1M=1mol/L

The unit factors are given below.

1M1mol/Land1mol/L1M

The unit factor to determine M from mol/L is given below.

1M1mol/L

Therefore, 0.5mol/L can be written as shown below.

Molarity=0.5mol/L×1M1mol/L=0.5M

Therefore, the molar concentration of HF solution is 0.5M.

Conclusion

The molar concentration of HF solution is 0.5M.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The molarity of hydrochloric acid in the solution is
Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive. (a) 52.3 mL of water is added to 23.3 mL of 0.123 M CaCl2 solution  M(b) 106 mL of water is added to 53.1 mL of 0.827 M KOH solution  M(c) 640. mL of water is added to 151 mL of 33.0 M NaCl solution  M(d) 365 mL of water is added to 724 mL of 1.92 M NaOH solution
The proccess of dissolving and hydrating a solute is

Chapter 15 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 15 - Prob. 2KTCh. 15 - Prob. 3KTCh. 15 - Prob. 4KTCh. 15 - Prob. 5KTCh. 15 - Prob. 6KTCh. 15 - Prob. 7KTCh. 15 - Prob. 8KTCh. 15 - Prob. 9KTCh. 15 - Prob. 10KTCh. 15 - Prob. 11KTCh. 15 - Prob. 12KTCh. 15 - Prob. 13KTCh. 15 - Prob. 14KTCh. 15 - Prob. 15KTCh. 15 - Prob. 16KTCh. 15 - Prob. 17KTCh. 15 - Prob. 18KTCh. 15 - Prob. 19KTCh. 15 - Prob. 20KTCh. 15 - Prob. 21KTCh. 15 - Prob. 22KTCh. 15 - Prob. 1ECh. 15 - Prob. 2ECh. 15 - Prob. 3ECh. 15 - Prob. 4ECh. 15 - Prob. 5ECh. 15 - Prob. 6ECh. 15 - Prob. 7ECh. 15 - Prob. 8ECh. 15 - Prob. 9ECh. 15 - Prob. 10ECh. 15 - Prob. 11ECh. 15 - Prob. 12ECh. 15 - Prob. 13ECh. 15 - Prob. 14ECh. 15 - Prob. 15ECh. 15 - Prob. 16ECh. 15 - Prob. 17ECh. 15 - Prob. 18ECh. 15 - Prob. 19ECh. 15 - Prob. 20ECh. 15 - Prob. 21ECh. 15 - Prob. 22ECh. 15 - Prob. 23ECh. 15 - Prob. 24ECh. 15 - Prob. 25ECh. 15 - Prob. 26ECh. 15 - Prob. 27ECh. 15 - Prob. 28ECh. 15 - Prob. 29ECh. 15 - Prob. 30ECh. 15 - Prob. 31ECh. 15 - Prob. 32ECh. 15 - Prob. 33ECh. 15 - Prob. 34ECh. 15 - Prob. 35ECh. 15 - Prob. 36ECh. 15 - Prob. 37ECh. 15 - Prob. 38ECh. 15 - Prob. 39ECh. 15 - Prob. 40ECh. 15 - Prob. 41ECh. 15 - Prob. 42ECh. 15 - Prob. 43ECh. 15 - Prob. 44ECh. 15 - Prob. 45ECh. 15 - Prob. 46ECh. 15 - Prob. 47ECh. 15 - Prob. 48ECh. 15 - Prob. 49ECh. 15 - Prob. 50ECh. 15 - Prob. 51ECh. 15 - Prob. 52ECh. 15 - Prob. 53ECh. 15 - Prob. 54ECh. 15 - Prob. 55ECh. 15 - Prob. 56ECh. 15 - Prob. 57ECh. 15 - Prob. 58ECh. 15 - Prob. 59ECh. 15 - Prob. 60ECh. 15 - Prob. 1STCh. 15 - Prob. 2STCh. 15 - Prob. 3STCh. 15 - Prob. 4STCh. 15 - Prob. 5STCh. 15 - Prob. 6STCh. 15 - Prob. 7STCh. 15 - Prob. 8STCh. 15 - Prob. 9STCh. 15 - Prob. 10STCh. 15 - Prob. 11STCh. 15 - Prob. 12STCh. 15 - Prob. 13STCh. 15 - Prob. 14STCh. 15 - Prob. 15STCh. 15 - Prob. 16ST
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Chemistry: An Active Learning Approa...
Chemistry
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Text book image
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY