Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Videos

Question
Book Icon
Chapter 13, Problem 82P

(a)

To determine

The molar mass of oleic acid.

(a)

Expert Solution
Check Mark

Answer to Problem 82P

The molar mass of oleic acid is 282.47 g/mol.

Explanation of Solution

The molecular formula of oleic acid is C18H34O2.

Write the expression for molar mass of oleic acid

Mm=[18(mC)+34(mH)+2(mO)]g/mol                                                           (I)

Here, Mm is the molar mass, mC is the standard atomic weight of carbon, mH is the standard atomic weight of hydrogen and mO is the standard atomic weight of oxygen.

Substitute 12.011 for mC, 1.00794 for mH and 15.9994 for mO in (I) to find Mm.

Mm=[18(12.011)+34(1.00794)+2(15.9994)]g/mol=282.47 g/mol

Thus, the molar mass of oleic acid is 282.47 g/mol.

(b)

To determine

The number of moles in one drop of oleic acid.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

The number of moles in one drop of oleic acid is 8.1×108mol.

Explanation of Solution

The mass of one drop of oleic acid is 2.3×105g and its volume is 2.6×105cm3.

Write the expression for number of moles

n=MMm                                                           (II)

Here, n is the number of moles and M is the given mass of oleic acid.

Substitute 2.3×105g for M and 282.47 g/mol for Mm in (II) to find n

n=2.3×105g282.47 g/mol=8.1×108mol

Thus, the number of moles is 8.1×108mol.

(c)

To determine

The side of the square of the base of the oleic acid.

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The side of the square is 5.3×108cm.

Explanation of Solution

One drop of oleic acid of volume 2.6×105cm3 forms a film of area 70.0cm2 with one molecule thickness when spread out on water. The height of one molecule is 7d. Here. d is the side of the square of the base of oleic acid.

Write the expression for volume

V=Ah                                                           (III)

Here, A is the area of the film, h is the height of the molecule and V is the volume.

Substitute 7d for h in (III)

V=A(7d)                                                             (IV)

Rearrange for d

d=V7A                                                             (V)

Substitute 70.0cm2 for A and 2.6×105cm3 for V in (V) to find d

d=2.6×105cm37(70.0cm2)=5.3×108cm

Thus, the side of the square is 5.3×108cm.

(d)

To determine

The number of molecules on the film.

(d)

Expert Solution
Check Mark

Answer to Problem 82P

The number of molecules on the film is 2.58×1016.

Explanation of Solution

Write the expression for number of molecules on the film

N=AA                                                           (VI)

Here, A is the total area of the film, N is the number of molecules on the film and A is the area of one molecule.

Substitute d2 for A in (VI)

N=Ad2                                                             (VII)

Substitute 70.0cm2 for A and 5.3×108cm for d in (VII) to find N

N=70.0cm2(5.3×108cm)2=2.58×1016

Thus, the number of molecules on the film is 2.58×1016.

(e)

To determine

An estimate of the Avogadro’s number.

(e)

Expert Solution
Check Mark

Answer to Problem 82P

The estimate of Avogadro’s number is 3.1×1023mol1.

Explanation of Solution

Write the expression for estimating Avogadro’s number using monolayer formation

NA=Nn                                                           (VIII)

Here, NA is the Avogadro’s number, N is the number of molecules in one drop and n is the number of moles in one drop.

Substitute 2.5×1016 for N and 8.1×108mol for n in (VIII) to find NA

NA=2.5×10168.1×108mol=3.1×1023mol1

Thus, the estimate of Avogadro’s number is 3.1×1023mol1.

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!

Chapter 13 Solutions

Physics

Ch. 13.6 - Prob. 13.7PPCh. 13.7 - Prob. 13.8PPCh. 13.8 - Prob. 13.9PPCh. 13 - Prob. 1CQCh. 13 - Prob. 2CQCh. 13 - Prob. 3CQCh. 13 - Prob. 4CQCh. 13 - Prob. 5CQCh. 13 - Prob. 6CQCh. 13 - Prob. 7CQCh. 13 - Prob. 8CQCh. 13 - Prob. 9CQCh. 13 - Prob. 10CQCh. 13 - Prob. 11CQCh. 13 - Prob. 12CQCh. 13 - Prob. 13CQCh. 13 - Prob. 14CQCh. 13 - Prob. 15CQCh. 13 - Prob. 16CQCh. 13 - Prob. 17CQCh. 13 - Prob. 18CQCh. 13 - Prob. 19CQCh. 13 - Prob. 20CQCh. 13 - Prob. 1MCQCh. 13 - Prob. 2MCQCh. 13 - Prob. 3MCQCh. 13 - Prob. 4MCQCh. 13 - Prob. 5MCQCh. 13 - Prob. 6MCQCh. 13 - Prob. 7MCQCh. 13 - Prob. 8MCQCh. 13 - Prob. 9MCQCh. 13 - Prob. 10MCQCh. 13 - Prob. 1PCh. 13 - Prob. 2PCh. 13 - Prob. 3PCh. 13 - Prob. 4PCh. 13 - Prob. 5PCh. 13 - Prob. 6PCh. 13 - Prob. 7PCh. 13 - Prob. 8PCh. 13 - Prob. 9PCh. 13 - Prob. 10PCh. 13 - Prob. 11PCh. 13 - Prob. 12PCh. 13 - Prob. 13PCh. 13 - Prob. 14PCh. 13 - Prob. 15PCh. 13 - Prob. 16PCh. 13 - Prob. 17PCh. 13 - Prob. 18PCh. 13 - Prob. 19PCh. 13 - Prob. 20PCh. 13 - Prob. 21PCh. 13 - 22. A copper washer is to be fit in place over a...Ch. 13 - 23. Repeat Problem 22, but now the copper washer...Ch. 13 - Prob. 24PCh. 13 - Prob. 25PCh. 13 - Prob. 26PCh. 13 - Prob. 27PCh. 13 - Prob. 28PCh. 13 - Prob. 29PCh. 13 - Prob. 30PCh. 13 - Prob. 31PCh. 13 - Prob. 32PCh. 13 - Prob. 33PCh. 13 - Prob. 34PCh. 13 - Prob. 35PCh. 13 - Prob. 36PCh. 13 - Prob. 37PCh. 13 - Prob. 38PCh. 13 - Prob. 39PCh. 13 - Prob. 40PCh. 13 - Prob. 41PCh. 13 - Prob. 42PCh. 13 - Prob. 43PCh. 13 - Prob. 44PCh. 13 - Prob. 45PCh. 13 - Prob. 46PCh. 13 - Prob. 47PCh. 13 - Prob. 48PCh. 13 - Prob. 49PCh. 13 - Prob. 50PCh. 13 - Prob. 51PCh. 13 - Prob. 52PCh. 13 - Prob. 53PCh. 13 - Prob. 54PCh. 13 - Prob. 55PCh. 13 - Prob. 56PCh. 13 - Prob. 57PCh. 13 - Prob. 58PCh. 13 - Prob. 59PCh. 13 - Prob. 60PCh. 13 - Prob. 61PCh. 13 - Prob. 62PCh. 13 - Prob. 63PCh. 13 - Prob. 64PCh. 13 - Prob. 65PCh. 13 - Prob. 66PCh. 13 - Prob. 67PCh. 13 - Prob. 68PCh. 13 - Prob. 69PCh. 13 - Prob. 70PCh. 13 - Prob. 71PCh. 13 - Prob. 72PCh. 13 - Prob. 73PCh. 13 - Prob. 74PCh. 13 - Prob. 75PCh. 13 - Prob. 76PCh. 13 - Prob. 77PCh. 13 - Prob. 78PCh. 13 - Prob. 79PCh. 13 - Prob. 80PCh. 13 - Prob. 81PCh. 13 - Prob. 82PCh. 13 - Prob. 83PCh. 13 - Prob. 84PCh. 13 - Prob. 85PCh. 13 - Prob. 86PCh. 13 - Prob. 87PCh. 13 - Prob. 88PCh. 13 - Prob. 89PCh. 13 - Prob. 90PCh. 13 - Prob. 91PCh. 13 - Prob. 92PCh. 13 - Prob. 93PCh. 13 - Prob. 94PCh. 13 - Prob. 95PCh. 13 - Prob. 96PCh. 13 - Prob. 97PCh. 13 - Prob. 98PCh. 13 - Prob. 99PCh. 13 - Prob. 100PCh. 13 - Prob. 101PCh. 13 - Prob. 102PCh. 13 - Prob. 103PCh. 13 - Prob. 104PCh. 13 - Prob. 105PCh. 13 - Prob. 106PCh. 13 - Prob. 107PCh. 13 - Prob. 108PCh. 13 - Prob. 109PCh. 13 - Prob. 110PCh. 13 - Prob. 111PCh. 13 - Prob. 112PCh. 13 - 113. A long, narrow steel rod of length 2.5000 m...Ch. 13 - Prob. 114PCh. 13 - Prob. 115PCh. 13 - Prob. 116PCh. 13 - Prob. 117PCh. 13 - Prob. 118PCh. 13 - Prob. 119PCh. 13 - Prob. 120P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
The Second Law of Thermodynamics: Heat Flow, Entropy, and Microstates; Author: Professor Dave Explains;https://www.youtube.com/watch?v=MrwW4w2nAMc;License: Standard YouTube License, CC-BY