Chapter 4, Problem 101QP

(a)

Interpretation Introduction

Interpretation:

The molarity of the acetic acid solution is to be determined.

Explanation of Solution

The molarity of the acetic acid solution is to be determined.

The number of moles is calculated from the following expression:

$n=\frac{m}{M_m}$ …… (1)

Here, $m$ is the mass of solute, $M_m$ is the molar mass of solute, and $n$ is the number of moles of solute.

The molarity of the solution is determined by the following expression:

$M=\frac{n}{V}$ …… (2)

Here, $n$ is the number of moles of solute, $V$ is the volume of the solution in liters, and $M$ is the molarity of the solution.

The molar mass of acetic acid is $60.05{\text{ gmol}}^{-1}$ .

Substitute $122\text{ g}$ for $m$ and $60.05{\text{ gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of ${\text{CH}}_{\text{3}}\text{COOH}$ as follows:

$\begin{array}{c}{n}_{{\text{CH}}_{\text{3}}\text{COOH}}=122\cancel{{\text{g CH}}_{\text{3}}\text{COOH}}\times \frac{{\text{1 mol CH}}_{\text{3}}\text{COOH}}{60.05\cancel{{\text{g CH}}_{\text{3}}\text{COOH}}}\\ =2.03{\text{ mol CH}}_{\text{3}}\text{COOH}\end{array}$

Substitute $2.03\text{ mol}$ for $n$ and $1.00\text{ L}$ for $V$ in equation (2) to determine the molarity of ${\text{CH}}_{\text{3}}\text{COOH}$ solution as follows:

$\begin{array}{c}{M}_{{\text{CH}}_{\text{3}}\text{COOH}}=\frac{2.03{\text{ mol CH}}_{\text{3}}\text{COOH}}{1.00\text{ L solution}}\\ =2.03{\text{ M CH}}_{\text{3}}\text{COOH}\end{array}$

Therefore, the molarity of ${\text{CH}}_{\text{3}}\text{COOH}$ solution is $2.03\text{ M}$ .

(b)

Interpretation Introduction

Interpretation:

The molarity of sucrose solution is to be determined.

Explanation of Solution

The molar mass of sucrose is $\text{342}{\text{.3 gmol}}^{-1}$ .

Substitute $185\text{ g}$ for $m$ and $\text{342}{\text{.3 gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ as follows:

$\begin{array}{c}{n}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}=185\cancel{{\text{g C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\times \frac{{\text{1 mol C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}{\text{342}\text{.3 }\cancel{{\text{g C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}}\\ =0.540{\text{ mol C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\end{array}$

Substitute $0.540\text{ mol}$ for $n$ and $1.00\text{ L}$ for $V$ in equation (2) to determine the molarity of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution as follows:

$\begin{array}{c}{M}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}=\frac{0.540{\text{ mol C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}{1.00\text{ L solution}}\\ =0.540{\text{ M C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\end{array}$

Therefore, the molarity of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution is $0.540\text{ M}$ .

(c)

Interpretation Introduction

Interpretation:

The molarity of hydrogen chloride solution is to be determined.

Explanation of Solution

The molar mass of hydrogen chloride is $\text{36}{\text{.46 gmol}}^{-1}$ .

Substitute $70.0\text{ g}$ for $m$ and $\text{36}{\text{.46 gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of $\text{HCl}$ as follows:

$\begin{array}{c}{n}_{\text{HCl}}=70.0\cancel{\text{g HCl}}\times \frac{\text{1 mol HCl}}{\text{36}\text{.46 }\cancel{\text{g HCl}}}\\ =1.92\text{ mol HCl}\end{array}$

Substitute $1.92\text{ mol}$ for $n$ and $\text{0}\text{.600 L}$ for $V$ in equation (2) to determine the molarity of $\text{HCl}$ solution as follows:

$\begin{array}{c}{M}_{\text{HCl}}=\frac{\text{1}\text{.92 mol HCl}}{\text{0}\text{.600 L solution}}\\ =3.20\text{ M HCl}\end{array}$

Therefore, the molarity of $\text{HCl}$ solution is $\text{3}\text{.20 M}$ .

(d)

Interpretation Introduction

Interpretation:

The molarity of the potassium hydroxide solution is to be determined.

Explanation of Solution

The molar mass of potassium hydroxide is $\text{56}{\text{.11 gmol}}^{-1}$ .

Substitute $45.0\text{ g}$ for $m$ and $\text{56}{\text{.11 gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of $\text{KOH}$ as follows:

$\begin{array}{c}{n}_{\text{KOH}}=45.0\cancel{\text{ g KOH}}\times \frac{\text{1 mol KOH}}{\text{56}\text{.11 }\cancel{\text{g KOH}}}\\ =0.802\text{ mol KOH}\end{array}$

The conversion of volume from milliliters to liters is given as follows:

$\text{Volume}\left(\text{L}\right)=x\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}$ …… (3)

Substitute $250.0\text{ mL}$ for $x$ in equation (3) to determine the volume of $\text{KOH}$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=250.0\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}\\ =0.2500\text{ L}\end{array}$

Substitute $0.802\text{ mol}$ for $n$ and $\text{0}\text{.2500 L}$ for $V$ in equation (2) to determine the molarity of $\text{KOH}$ solution as follows:

$\begin{array}{c}{M}_{\text{KOH}}=\frac{0.802\text{ mol KOH}}{\text{0}\text{.2500 L solution}}\\ =3.21\text{ M KOH}\end{array}$

Therefore, the molarity of $\text{KOH}$ solution is $\text{3}\text{.21 M}$ .