Chapter 14, Problem 40PE

(a)

Interpretation Introduction

Interpretation:

Molality of solution prepared by dissolution of $\text{1}{\text{.25 mol CaCl}}_2$ in $750.0\text{ g}$ of water has to be determined.

Concept Introduction:

Molality is amount of solute present in one kilogram of solvent. Expression to calculate 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 ${\text{CaCl}}_2$ is as follows:

  ${\text{Molality of CaCl}}_2=\frac{{\text{Moles of CaCl}}_2}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (1)

Substitute $\text{1}\text{.25 mol}$ for moles of ${\text{CaCl}}_2$ and $750.0\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (1).

  $\begin{array}{c}{\text{Molality of CaCl}}_2=\left(\frac{\text{1}\text{.25 mol}}{750.0\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =1.67m\end{array}$

Hence, molality of required solution is $1.67m$.

(b)

Interpretation Introduction

Interpretation:

Molality of solution prepared by dissolution of $2.5{\text{ g C}}_6{\text{H}}_{12}{\text{O}}_6$ in $\text{525 g}$ of water has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression for moles of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is as follows:

  ${\text{Moles of C}}_6{\text{H}}_{12}{\text{O}}_6=\frac{{\text{Mass of C}}_6{\text{H}}_{12}{\text{O}}_6}{{\text{Molar mass of C}}_6{\text{H}}_{12}{\text{O}}_6}$        (2)

Substitute $\text{2}\text{.5 g}$ for mass and $180.1\text{ g/mol}$ for molar mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ in equation (2).

  $\begin{array}{c}{\text{Moles of C}}_6{\text{H}}_{12}{\text{O}}_6=\frac{\text{2}\text{.5 g}}{180.1\text{ g/mol}}\\ =0.0139\text{ mol}\end{array}$

Expression for molality of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is as follows:

  ${\text{Molality of C}}_6{\text{H}}_{12}{\text{O}}_6=\frac{{\text{Moles of C}}_6{\text{H}}_{12}{\text{O}}_6}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (3)

Substitute $0.0139\text{ mol}$ for moles of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ and $\text{525 g}$ for mass of ${\text{H}}_2\text{O}$ in equation (3).

  $\begin{array}{c}{\text{Molality of C}}_6{\text{H}}_{12}{\text{O}}_6=\left(\frac{0.0139\text{ mol}}{\text{525 g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =0.0265m\end{array}$

Hence, molality of required solution is $0.0265m$.

(c)

Interpretation Introduction

Interpretation:

Molality of solution prepared by dissolution of $\text{17}\text{.5 mL }{\left({\text{CH}}_3\right)}_2\text{CHOH}$ in $\text{35}{\text{.5 mL H}}_2\text{O}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression for mass of ${\left({\text{CH}}_3\right)}_2\text{CHOH}$ is as follows:

  $\text{Mass of }{\left({\text{CH}}_3\right)}_2\text{CHOH}=\left[\begin{array}{c}\left(\text{Density of }{\left({\text{CH}}_3\right)}_2\text{CHOH}\right)\\ \left(\text{Volume of }{\left({\text{CH}}_3\right)}_2\text{CHOH}\right)\end{array}\right]$        (4)

Substitute $\text{0}\text{.785 g/mL}$ for density and $\text{17}\text{.5 mL}$ for volume of ${\left({\text{CH}}_3\right)}_2\text{CHOH}$ in equation (4).

  $\begin{array}{c}\text{Mass of }{\left({\text{CH}}_3\right)}_2\text{CHOH}=\left(\text{0}\text{.785 g/mL}\right)\left(\text{17}\text{.5 mL}\right)\\ =13.74\text{ g}\end{array}$

Expression for mass of ${\text{H}}_2\text{O}$ is as follows:

  ${\text{Mass of H}}_2\text{O}=\left({\text{Density of H}}_2\text{O}\right)\left({\text{Volume of H}}_2\text{O}\right)$        (5)

Substitute $\text{1}\text{.00 g/mL}$ for density and $\text{35}\text{.5 mL}$ for volume of ${\text{H}}_2\text{O}$ in equation (5).

  $\begin{array}{c}{\text{Mass of H}}_2\text{O}=\left(\text{1}\text{.00 g/mL}\right)\left(\text{35}\text{.5 mL}\right)\\ =35.5\text{ g}\end{array}$

Expression for moles of ${\left({\text{CH}}_3\right)}_2\text{CHOH}$ is as follows:

  $\text{Moles of }{\left({\text{CH}}_3\right)}_2\text{CHOH}=\frac{\text{Mass of }{\left({\text{CH}}_3\right)}_2\text{CHOH}}{\text{Molar mass of }{\left({\text{CH}}_3\right)}_2\text{CHOH}}$        (6)

Substitute $\text{13}\text{.74 g}$ for mass and $60.1\text{ g/mol}$ for molar mass of ${\left({\text{CH}}_3\right)}_2\text{CHOH}$ in equation (6).

  $\begin{array}{c}\text{Moles of }{\left({\text{CH}}_3\right)}_2\text{CHOH}=\frac{\text{13}\text{.74 g}}{60.1\text{ g/mol}}\\ =0.229\text{ mol}\end{array}$

Expression for molality of ${\left({\text{CH}}_3\right)}_2\text{CHOH}$ is as follows:

  $\text{Molality of }{\left({\text{CH}}_3\right)}_2\text{CHOH}=\frac{\text{Moles of }{\left({\text{CH}}_3\right)}_2\text{CHOH}}{\text{Mass}\left(\text{kg}\right){\text{ of H}}_2\text{O}}$        (7)

Substitute $0.229\text{ mol}$ for moles of ${\left({\text{CH}}_3\right)}_2\text{CHOH}$ and $35.5\text{ g}$ for mass of ${\text{H}}_2\text{O}$ in equation (7).

  $\begin{array}{c}\text{Molality of }{\left({\text{CH}}_3\right)}_2\text{CHOH}=\left(\frac{0.229\text{ mol}}{35.5\text{ g}}\right)\left(\frac{1\text{ g}}{{10}^{-3}\text{ kg}}\right)\\ =6.45m\end{array}$

Hence, molality of required solution is $6.45m$.