PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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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.

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Chapter 13 Solutions

PHYSICS

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