Chapter 14, Problem 34RQ

(a)

Interpretation Introduction

Interpretation:

Whether $100\text{ g}$ of sucrose or $100\text{ g}$ of ethyl alcohol is effective to lower freezing point of $500\text{ g}$ of water has to be determined.

Concept Introduction:

Colligative properties are dependent on amount of solute particles and not on their natures. These are caused due to addition of nonvolatile solute in any solution. Below mentioned are colligative properties.

1. Freezing point depression

2. Boiling point elevation

3. Vapor pressure lowering

4. Osmotic pressure

Explanation of Solution

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

  ${\text{Moles of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\frac{{\text{Mass of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}{{\text{Molar mass of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}$        (1)

Substitute $100\text{ g}$ for mass and $342.3\text{ g/mol}$ for molar mass of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ in equation (1).

  $\begin{array}{c}{\text{Moles of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\frac{100\text{ g}}{342.3\text{ g/mol}}\\ =0.292\text{ mol}\end{array}$

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

  ${\text{Molality of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\frac{{\text{Moles of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (2)

Substitute $0.292\text{ mol}$ for moles of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ and $500\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (2).

  $\begin{array}{c}{\text{Molality of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\left(\frac{0.292\text{ mol}}{500\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =0.584m\end{array}$

Formula for moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ is as follows:

  ${\text{Moles of C}}_2{\text{H}}_5\text{OH}=\frac{{\text{Mass of C}}_2{\text{H}}_5\text{OH}}{{\text{Molar mass of C}}_2{\text{H}}_5\text{OH}}$        (3)

Substitute $100\text{ g}$ for mass and $46.07\text{ g/mol}$ for molar mass of ${\text{C}}_2{\text{H}}_5\text{OH}$ in equation (3).

  $\begin{array}{c}{\text{Moles of C}}_2{\text{H}}_5\text{OH}=\frac{100\text{ g}}{\text{46}\text{.07 g/mol}}\\ =2.17\text{ mol}\end{array}$

Expression for molality of ${\text{C}}_2{\text{H}}_5\text{OH}$ is as follows:

  ${\text{Molality of C}}_2{\text{H}}_5\text{OH}=\frac{{\text{Moles of C}}_2{\text{H}}_5\text{OH}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (4)

Substitute $2.17\text{ mol}$ for moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ and $500\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (4).

  $\begin{array}{c}{\text{Molality of C}}_2{\text{H}}_5\text{OH}=\left(\frac{\text{2}\text{.17 mol}}{500\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =4.34m\end{array}$

Depression in freezing point is directly related to molality. Since molality of ${\text{C}}_2{\text{H}}_5\text{OH}$ is more than that of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$, $100\text{ g}$ of ethyl alcohol is more effective to lower freezing point of $500\text{ g}$ of water.

(b)

Interpretation Introduction

Interpretation:

Whether $100\text{ g}$ of sucrose or $20.0\text{ g}$ of ethyl alcohol is effective to lower freezing point of $500\text{ g}$ of water has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

  ${\text{Moles of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\frac{{\text{Mass of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}{{\text{Molar mass of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}$        (1)

Substitute $100\text{ g}$ for mass and $342.3\text{ g/mol}$ for molar mass of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ in equation (1).

  $\begin{array}{c}{\text{Moles of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\frac{100\text{ g}}{342.3\text{ g/mol}}\\ =0.292\text{ mol}\end{array}$

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

  ${\text{Molality of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\frac{{\text{Moles of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (2)

Substitute $0.292\text{ mol}$ for moles of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ and $500\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (2).

  $\begin{array}{c}{\text{Molality of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\left(\frac{0.292\text{ mol}}{500\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =0.584m\end{array}$

Formula for moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ is as follows:

  ${\text{Moles of C}}_2{\text{H}}_5\text{OH}=\frac{{\text{Mass of C}}_2{\text{H}}_5\text{OH}}{{\text{Molar mass of C}}_2{\text{H}}_5\text{OH}}$        (3)

Substitute $20.0\text{ g}$ for mass and $46.07\text{ g/mol}$ for molar mass of ${\text{C}}_2{\text{H}}_5\text{OH}$ in equation (3).

  $\begin{array}{c}{\text{Moles of C}}_2{\text{H}}_5\text{OH}=\frac{20.0\text{ g}}{\text{46}\text{.07 g/mol}}\\ =0.434\text{ mol}\end{array}$

Expression for molality of ${\text{C}}_2{\text{H}}_5\text{OH}$ is as follows:

  ${\text{Molality of C}}_2{\text{H}}_5\text{OH}=\frac{{\text{Moles of C}}_2{\text{H}}_5\text{OH}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (4)

Substitute $0.434\text{ mol}$ for moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ and $500\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (4).

  $\begin{array}{c}{\text{Molality of C}}_2{\text{H}}_5\text{OH}=\left(\frac{0.434\text{ mol}}{500\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =0.868m\end{array}$

Depression in freezing point is directly related to molality. Since molality of ${\text{C}}_2{\text{H}}_5\text{OH}$ is more than that of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$, $20.0\text{ g}$ of ethyl alcohol is more effective to lower freezing point of $500\text{ g}$ of water.

(c)

Interpretation Introduction

Interpretation:

Whether $20.0\text{ g}$ of ethyl alcohol or $20.0\text{ g}$ of methyl alcohol is effective to lower freezing point of $500\text{ g}$ of water has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Formula for moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ is as follows:

  ${\text{Moles of C}}_2{\text{H}}_5\text{OH}=\frac{{\text{Mass of C}}_2{\text{H}}_5\text{OH}}{{\text{Molar mass of C}}_2{\text{H}}_5\text{OH}}$        (3)

Substitute $20.0\text{ g}$ for mass and $46.07\text{ g/mol}$ for molar mass of ${\text{C}}_2{\text{H}}_5\text{OH}$ in equation (3).

  $\begin{array}{c}{\text{Moles of C}}_2{\text{H}}_5\text{OH}=\frac{20.0\text{ g}}{\text{46}\text{.07 g/mol}}\\ =0.434\text{ mol}\end{array}$

Expression for molality of ${\text{C}}_2{\text{H}}_5\text{OH}$ is as follows:

  ${\text{Molality of C}}_2{\text{H}}_5\text{OH}=\frac{{\text{Moles of C}}_2{\text{H}}_5\text{OH}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (4)

Substitute $0.434\text{ mol}$ for moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ and $500\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (4).

  $\begin{array}{c}{\text{Molality of C}}_2{\text{H}}_5\text{OH}=\left(\frac{0.434\text{ mol}}{500\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =0.868m\end{array}$

Formula for moles of ${\text{CH}}_3\text{OH}$ is as follows:

  ${\text{Moles of CH}}_3\text{OH}=\frac{{\text{Mass of CH}}_3\text{OH}}{{\text{Molar mass of CH}}_3\text{OH}}$        (5)

Substitute $20.0\text{ g}$ for mass and $\text{32}\text{.04 g/mol}$ for molar mass of ${\text{CH}}_3\text{OH}$ in equation (5).

  $\begin{array}{c}{\text{Moles of CH}}_3\text{OH}=\frac{20.0\text{ g}}{\text{32}\text{.04 g/mol}}\\ =0.624\text{ mol}\end{array}$

Expression for molality of ${\text{CH}}_3\text{OH}$ is as follows:

  ${\text{Molality of CH}}_3\text{OH}=\frac{{\text{Moles of CH}}_3\text{OH}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (6)

Substitute $0.624\text{ mol}$ for moles of ${\text{CH}}_3\text{OH}$ and $500\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (4).

  $\begin{array}{c}{\text{Molality of CH}}_3\text{OH}=\left(\frac{0.624\text{ mol}}{500\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =1.248m\end{array}$

Depression in freezing point is directly related to molality. Since molality of ${\text{CH}}_3\text{OH}$ is more than that of ${\text{C}}_2{\text{H}}_5\text{OH}$, $20.0\text{ g}$ of methyl alcohol is more effective to lower freezing point of $500\text{ g}$ of water.