Chapter 13, Problem 82P

(a)

To determine

The molar mass of oleic acid.

Answer to Problem 82P

The molar mass of oleic acid is $282.47\text{ g/mol}$.

Explanation of Solution

The molecular formula of oleic acid is ${\text{C}}_{18}{\text{H}}_{34}{\text{O}}_2$.

Write the expression for molar mass of oleic acid

$M_{\text{m}}=\left[18\left(m_{\text{C}}\right)+34\left(m_{\text{H}}\right)+2\left(m_{\text{O}}\right)\right]\text{g/mol}$                                                           (I)

Here, $M_{\text{m}}$ is the molar mass, $m_{\text{C}}$ is the standard atomic weight of carbon, $m_{\text{H}}$ is the standard atomic weight of hydrogen and $m_{\text{O}}$ is the standard atomic weight of oxygen.

Substitute $12.011$ for $m_{\text{C}}$, $1.00794$ for $m_{\text{H}}$ and $15.9994$ for $m_{\text{O}}$ in (I) to find $M_{\text{m}}$.

$\begin{array}{c}{M}_{\text{m}}=\left[18\left(12.011\right)+34\left(1.00794\right)+2\left(15.9994\right)\right]\text{g/mol}\\ =282.47\text{ g/mol}\end{array}$

Thus, the molar mass of oleic acid is $282.47\text{ g/mol}$.

(b)

To determine

The number of moles in one drop of oleic acid.

Answer to Problem 82P

The number of moles in one drop of oleic acid is $8.1\times {10}^{-8}\text{mol}$.

Explanation of Solution

The mass of one drop of oleic acid is $2.3\times {10}^{-5}\text{}\text{g}$ and its volume is $2.6\times {10}^{-5}{\text{cm}}^3$.

Write the expression for number of moles

$n=\frac{M}{M_{\text{m}}}$                                                           (II)

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

Substitute $2.3\times {10}^{-5}\text{}\text{g}$ for $M$ and $282.47\text{ g/mol}$ for $M_{\text{m}}$ in (II) to find $n$

$\begin{array}{c}n=\frac{2.3\times {10}^{-5}\text{}\text{g}}{282.47\text{ g/mol}}\\ =8.1\times {10}^{-8}\text{mol}\end{array}$

Thus, the number of moles is $8.1\times {10}^{-8}\text{mol}$.

(c)

To determine

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

Answer to Problem 82P

The side of the square is $5.3\times {10}^{-8}\text{cm}$.

Explanation of Solution

One drop of oleic acid of volume $2.6\times {10}^{-5}{\text{cm}}^3$ forms a film of area $70.0{\text{cm}}^2$ 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)

$V=A\left(7d\right)$                                                             (IV)

Rearrange for $d$

$d=\frac{V}{7A}$                                                             (V)

Substitute $70.0{\text{cm}}^2$ for $A$ and $2.6\times {10}^{-5}{\text{cm}}^3$ for $V$ in (V) to find $d$

$\begin{array}{c}d=\frac{2.6\times {10}^{-5}{\text{cm}}^3}{7\left(70.0{\text{cm}}^2\right)}\\ =5.3\times {10}^{-8}\text{cm}\end{array}$

Thus, the side of the square is $5.3\times {10}^{-8}\text{cm}$.

(d)

To determine

The number of molecules on the film.

Answer to Problem 82P

The number of molecules on the film is $2.58\times {10}^{16}$.

Explanation of Solution

Write the expression for number of molecules on the film

$N=\frac{A}{A^{\prime }}$                                                           (VI)

Here, $A$ is the total area of the film, $N$ is the number of molecules on the film and $A^{\prime }$ is the area of one molecule.

Substitute $d^2$ for $A^{\prime }$ in (VI)

$N=\frac{A}{d^2}$                                                             (VII)

Substitute $70.0{\text{cm}}^2$ for $A$ and $5.3\times {10}^{-8}\text{cm}$ for $d$ in (VII) to find $N$

$\begin{array}{c}N=\frac{70.0{\text{cm}}^2}{{\left(5.3\times {10}^{-8}\text{cm}\right)}^2}\\ =2.58\times {10}^{16}\end{array}$

Thus, the number of molecules on the film is $2.58\times {10}^{16}$.

(e)

To determine

An estimate of the Avogadro’s number.

Answer to Problem 82P

The estimate of Avogadro’s number is $3.1\times {10}^{23}{\text{mol}}^{-1}$.

Explanation of Solution

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

$N_{\text{A}}=\frac{N}{n}$                                                           (VIII)

Here, $N_{\text{A}}$ 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 $2.5\times {10}^{16}$ for $N$ and $8.1\times {10}^{-8}\text{mol}$ for $n$ in (VIII) to find $N_{\text{A}}$

$\begin{array}{c}{N}_{\text{A}}=\frac{2.5\times {10}^{16}}{8.1\times {10}^{-8}\text{mol}}\\ =3.1\times {10}^{23}{\text{mol}}^{-1}\end{array}$

Thus, the estimate of Avogadro’s number is $3.1\times {10}^{23}{\text{mol}}^{-1}$.