Chapter 7, Problem 7.55E

Determine how ideal the following solutions are by calculating the mole fraction of solute in each solution, and comparing that to the expected mole fractions. All data are for $25.0°\text{C}$.

(a) $14.09$ weight percent of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$, MP of ${\text{I}}_{\text{2}}$ is $112.9°\text{C}$ (sublimes), and ${\Delta }_{\text{fus}}H=15.27\text{kJ}/\text{mol}$

(b) $2.72$ weight percent of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$, MP of ${\text{I}}_{\text{2}}$ is $112.9°\text{C}$ (sublimes), and ${\Delta }_{\text{fus}}H=15.27\text{kJ}/\text{mol}$

(c) $20.57$ weight percent of para-dichlorobenzene, ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$, in hexane, MP of ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ is $52.7°\text{C}$ and ${\Delta }_{\text{fus}}H=17.15\text{kJ}/\text{mol}$

Interpretation Introduction

(a)

Interpretation:

Whether the solution of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is ideal by calculation of mole fraction and its comparison with expected value is to be stated.

Concept introduction:

The mole fraction is given by the formula given below.

$\ln x_{\text{solute}}=-\frac{{\Delta }_{\text{fus}}H}{R}\left(\frac{1}{T}-\frac{1}{T_{\text{MP}}}\right)$

In the above expression, ${\Delta }_{\text{fus}}H$ is the enthalpy of fusion, $x_{\text{solute}}$ is the mole fraction of solute, $T_{\text{MP}}$ is the melting point, $T$ is the temperature at which reaction takes place and $R$ is the gas constant.

Answer to Problem 7.55E

The solution of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ has a value of $x_{\text{iodine}}=0.246$ which is higher than expected value of $\left(0.0478\right)$ due to non-ideal behavior of solution.

Explanation of Solution

The mole fraction of iodine is calculated as follows.

$\begin{array}{rll}\ln x_{\text{iodine}}& =& -\frac{{\Delta }_{\text{fus}}H}{R}\left(\frac{1}{T}-\frac{1}{T_{\text{MP}}}\right)\\ x_{\text{iodine}}& =& e^{\left(-\frac{15.27}{0.008314}\left(\frac{1}{298}-\frac{1}{385.9}\right)\right)}\\ x_{\text{iodine}}& =& 0.246\end{array}$

For $14.09$, weight percent of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$, number of moles of iodine expected are calculated as follows.

$x_{\text{iodine}}=\frac{n_{\text{iodine}}}{n_{\text{iodine}}+n_{\text{solvent}}}$

$\begin{array}{rll}{n}_{\text{iodine}}& =& \frac{14.09}{253.8}\\ & =& 0.055\text{mol}\end{array}$

$\begin{array}{rll}{n}_{\text{solvent}}& =& \frac{\left(100-14.09\right)}{78.11}\\ & =& 1.10\text{mol}\end{array}$

The mole fraction of iodine is calculated as shown below.

$\begin{array}{rll}{x}_{\text{iodine}}& =& \frac{n_{\text{iodine}}}{n_{\text{iodine}}+n_{\text{solvent}}}\\ & =& \frac{0.055}{0.055+1.10}\\ & =& 0.0478\end{array}$

Thus, the mole fraction is approximately five times higher than expected due to non-ideal behavior.

Conclusion

The mole fraction $\left(x_{\text{iodine}}=0.246\right)$ is approximately five times higher than expected $\left(0.0478\right)$ due to non-ideal behavior of the solution.

Interpretation Introduction

(b)

Interpretation:

Whether the solution of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ is ideal by calculation of mole fraction and its comparison with expected value is to be stated.

Concept introduction:

The mole fraction is given by the formula given below.

$\ln x_{\text{solute}}=-\frac{{\Delta }_{\text{fus}}H}{R}\left(\frac{1}{T}-\frac{1}{T_{\text{MP}}}\right)$

In the above expression, ${\Delta }_{\text{fus}}H$ is the enthalpy of fusion, $x_{\text{solute}}$ is the mole fraction of solute, $T_{\text{MP}}$ is the melting point, $T$ is the temperature at which reaction takes place and $R$ is the gas constant.

Answer to Problem 7.55E

The solution of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ has a value of $x_{\text{iodine}}=0.246$ which is higher than expected value of $0.00914$ due to non-ideal behavior of solution.

Explanation of Solution

The mole fraction of iodine is calculated as follows.

$\begin{array}{rll}\ln x_{\text{iodine}}& =& -\frac{{\Delta }_{\text{fus}}H}{R}\left(\frac{1}{T}-\frac{1}{T_{\text{MP}}}\right)\\ x_{\text{iodine}}& =& e^{\left(-\frac{15.27}{0.008314}\left(\frac{1}{298}-\frac{1}{385.9}\right)\right)}\\ x_{\text{iodine}}& =& 0.246\end{array}$

For $2.72$ weight percent of ${\text{I}}_{\text{2}}$ in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$, number of moles of iodine expected are calculated as follows.

$x_{\text{iodine}}=\frac{n_{\text{iodine}}}{n_{\text{iodine}}+n_{\text{solvent}}}$

$\begin{array}{rll}{n}_{\text{iodine}}& =& \frac{2.72}{253.8}\\ & =& 0.0107\text{mol}\end{array}$

$\begin{array}{rll}{n}_{\text{solvent}}& =& \frac{\left(100-2.72\right)}{84.46}\\ & =& 1.16\text{mol}\end{array}$

The mole fraction of iodine is calculated as follows.

$\begin{array}{rll}{x}_{\text{iodine}}& =& \frac{n_{\text{iodine}}}{n_{\text{iodine}}+n_{\text{solvent}}}\\ & =& \frac{0.0107}{0.0107+1.16}\\ & =& 0.00914\end{array}$

Thus, the mole fraction is approximately twenty seven times higher than expected value due to non-ideal behavior.

Conclusion

The mole fraction $\left(x_{\text{iodine}}=0.246\right)$ is approximately twenty seven times higher than expected value $\left(0.00914\right)$ due to non-ideal behavior of the solution.

Interpretation Introduction

(c)

Interpretation:

Whether the solution of para-dichlorobenzene, ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ in hexane is ideal by calculation of mole fraction and its comparison with expected value is to bestated.

Concept introduction:

The mole fraction is given by the formula given below.

$\ln x_{\text{solute}}=-\frac{{\Delta }_{\text{fus}}H}{R}\left(\frac{1}{T}-\frac{1}{T_{\text{MP}}}\right)$

In the above expression, ${\Delta }_{\text{fus}}H$ is the enthalpy of fusion, $x_{\text{solute}}$ is the mole fraction of solute, $T_{\text{MP}}$ is the melting point, $T$ is the temperature at which reaction takes place and $R$ is the gas constant.

Answer to Problem 7.55E

The solution of para-dichlorobenzene, ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ in hexane has a value of $x_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}=0.55$ which is higher than expected value $\left(0.132\right)$ due to non-ideal behavior of solution.

Explanation of Solution

The mole fraction of ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ is calculated as follows.

$\begin{array}{rll}\ln x_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}& =& -\frac{{\Delta }_{\text{fus}}H}{R}\left(\frac{1}{T}-\frac{1}{T_{\text{MP}}}\right)\\ x_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}& =& e^{\left(-\frac{17.15}{0.008314}\left(\frac{1}{298}-\frac{1}{325.7}\right)\right)}\\ x_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}& =& 0.55\end{array}$

For $20.57$ weight percent of ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ in hexane, number of moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ expected are calculated as follows.

$x_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}=\frac{n_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}}{n_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}+n_{\text{solvent}}}$

$\begin{array}{rll}{n}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}& =& \frac{20.57}{147.01}\\ & =& 0.140\text{mol}\end{array}$

$\begin{array}{rll}{n}_{\text{solvent}}& =& \frac{\left(100-20.57\right)}{86.18}\\ & =& 0.922\text{mol}\end{array}$

The mole fraction of para-dichlorobenzene is calculated as follows.

$\begin{array}{rll}{x}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}& =& \frac{n_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}}{n_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}+n_{\text{solvent}}}\\ & =& \frac{0.140}{0.140+0.922}\\ & =& 0.132\end{array}$

Thus, the mole fraction is approximately four times higher than expected value due to non-ideal behavior.

Conclusion

The mole fraction $\left(x_{{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}=0.55\right)$ is approximately four times higher than expected value $\left(0.132\right)$ due to non-ideal behavior of the solution.