Chapter 14, Problem 63AE

(a)

Interpretation Introduction

Interpretation:

Grams of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ present in $1.0\text{ L}$ of syrup have to be determined.

Concept Introduction:

Density of substance is ratio of mass per unit volume. Expression for density of substance is as follows:

  $\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Explanation of Solution

Concentration of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution is $15\text{ \%}$. This indicates that $15\text{ g}$ of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is present in $100\text{ g}$ of solution.

Expression for density of solution is as follows:

  $d=\frac{M}{V}$        (1)

Here,

$d$ is density of solution.

$M$ is mass of solution.

$V$ is volume of solution.

Rearrange equation (1) for $V$.

  $V=\frac{M}{d}$        (2)

Substitute $100\text{ g}$ for $M$ and $1.06\text{ g/mL}$ for $d$ in equation (2).

  $\begin{array}{c}V=\frac{100\text{ g}}{1.06\text{ g/mL}}\\ =94.34\text{ mL}\end{array}$

Since $94.34\text{ mL}$ of solution contains $15\text{ g}$ of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$, mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ present in $1.00\text{ L}$ of solution is calculated as follows:

  $\begin{array}{c}{\text{Mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\left(\frac{15\text{ g}}{\text{94}\text{.34 mL}}\right)\left(1.00\text{ L}\right)\left(\frac{{10}^3\text{ mL}}{1\text{ L}}\right)\\ =158.9\text{ g}\end{array}$

Hence, mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is $158.9\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

Molarity of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution has to be determined.

Concept Introduction:

Molarity is amount of solute per $1\text{ L}$ of solution. Expression to calculate molarity is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$

Explanation of Solution

Concentration of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution is $15\text{ \%}$. This indicates that $15\text{ g}$ of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is present in $100\text{ g}$ of solution.

Expression to calculate molarity of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is as follows:

  ${\text{Molarity of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\frac{{\text{Moles of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}{\text{Volume }\left(\text{L}\right){\text{ of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{ solution}}$        (3)

Expression for moles of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is as follows:

  ${\text{Moles of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\frac{{\text{Mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}{{\text{Molar mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}$        (4)

Substitute $15\text{ g}$ for mass and $342.3\text{ g/mol}$ for molar mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in equation (4).

  $\begin{array}{c}{\text{Moles of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\frac{15\text{ g}}{342.3\text{ g/mol}}\\ =0.0438\text{ mol}\end{array}$

Substitute $0.0438\text{ mol}$ for moles and $94.34\text{ mL}$ for volume of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution in equation (3).

  $\begin{array}{c}{\text{Molarity of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\left(\frac{0.0438\text{ mol}}{94.34\text{ mL}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\\ =0.464M\end{array}$

Hence, molarity of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is $0.464M$.

(c)

Interpretation Introduction

Interpretation:

Molality of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ solution has to be determined.

Concept Introduction:

Molality is amount of solute present in one kilogram of solvent. Expression for molality of solution is as follows:

  $\text{Molality of solution}=\frac{\text{Moles of solute}}{\text{Mass }\left(\text{kg}\right)\text{of solvent}}$

Explanation of Solution

Expression for molality of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is as follows:

  ${\text{Molality of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\frac{{\text{Moles of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (5)

Expression for mass of solution is as follows:

  $\text{Mass of solution}={\text{Mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}+{\text{Mass of H}}_2\text{O}$        (6)

Rearrange equation (6) for mass of ${\text{H}}_2\text{O}$.

  ${\text{Mass of H}}_2\text{O}=\text{Mass of solution}-{\text{Mass of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$        (7)

Substitute $100\text{ g}$ for mass of solution and $15\text{ g}$ for mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in equation (7).

  $\begin{array}{c}{\text{Mass of H}}_2\text{O}=100\text{ g}-15\text{ g}\\ =85\text{ g}\end{array}$

Substitute $0.0438\text{ mol}$ for moles of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and $\text{85 g}$ for mass of ${\text{H}}_2\text{O}$ in equation (5).

  $\begin{array}{c}{\text{Molality of C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}=\left(\frac{0.0438\text{ mol}}{\text{85 g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =0.515m\end{array}$

Hence, molality of required solution is $0.515m$.