Questions: The boiling point of chloroform, CHCl3, is 61.70°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in chloroform is testosterone. A student dissolves 11.39 grams of testosterone, C19H28O2 (288.4 g/mol), in 297.3 grams of chloroform. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb (°C/m) Kf (°C/m) ------------ Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The boiling point of the solution is °C.

The boiling point of chloroform, CHCl3, is 61.70°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in chloroform is testosterone. A student dissolves 11.39 grams of testosterone, C19H28O2 (288.4 g/mol), in 297.3 grams of chloroform. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb (°C/m) Kf (°C/m) ------------ Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The boiling point of the solution is °C.
Transcript text: The boiling point of chloroform, $\mathrm{CHCl}_{3}$, is $61.70^{\circ} \mathrm{C}$ at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in chloroform is testosterone. A student dissolves 11.39 grams of testosterone, $\mathrm{C}_{19} \mathrm{H}_{28} \mathrm{O}_{2}$ ( $288.4 \mathrm{~g} / \mathrm{mol}$ ), in 297.3 grams of chloroform. Use the table of boiling and freezing point constants to answer the questions below. \begin{tabular}{|c|c|c|c|} \hline Solvent & Formula & \begin{tabular}{c} $K_{\mathrm{b}}$ \\ $\left({ }^{\circ} \mathbf{C} / \mathbf{m}\right)$ \end{tabular} & \begin{tabular}{c} $K_{\mathrm{f}}$ \\ $\left({ }^{\circ} \mathbf{C} / \mathbf{m}\right)$ \end{tabular} \\ \hline Water & $\mathrm{H}_{2} \mathrm{O}$ & 0.512 & 1.86 \\ \hline Ethanol & $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ & 1.22 & 1.99 \\ \hline Chloroform & $\mathrm{CHCl}_{3}$ & 3.67 & \\ \hline Benzene & $\mathrm{C}_{6} \mathrm{H}_{6}$ & 2.53 & 5.12 \\ \hline \begin{tabular}{c} Diethyl \\ ether \end{tabular} & $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}$ & 2.02 & \\ \hline \end{tabular} The molality of the solution is $\square$ m . The boiling point of the solution is $\square$ ${ }^{\circ} \mathrm{C}$.
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Solution

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Solution Steps

Step 1: Calculate the Molality of the Solution

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the moles of testosterone:

\[ \text{Moles of testosterone} = \frac{\text{mass of testosterone}}{\text{molar mass of testosterone}} = \frac{11.39 \, \text{g}}{288.4 \, \text{g/mol}} = 0.0395 \, \text{mol} \]

Next, convert the mass of chloroform to kilograms:

\[ \text{Mass of chloroform in kg} = \frac{297.3 \, \text{g}}{1000} = 0.2973 \, \text{kg} \]

Now, calculate the molality:

\[ m = \frac{\text{moles of testosterone}}{\text{mass of chloroform in kg}} = \frac{0.0395 \, \text{mol}}{0.2973 \, \text{kg}} = 0.1329 \, \text{m} \]

Step 2: Calculate the Boiling Point Elevation

The boiling point elevation (\(\Delta T_b\)) can be calculated using the formula:

\[ \Delta T_b = K_b \times m \]

where \(K_b\) for chloroform is \(3.67 \, ^\circ \text{C/m}\). Substituting the values:

\[ \Delta T_b = 3.67 \, ^\circ \text{C/m} \times 0.1329 \, \text{m} = 0.4875 \, ^\circ \text{C} \]

Step 3: Calculate the Boiling Point of the Solution

The boiling point of the solution is the sum of the boiling point of the pure solvent and the boiling point elevation:

\[ \text{Boiling point of solution} = 61.70 \, ^\circ \text{C} + 0.4875 \, ^\circ \text{C} = 62.1875 \, ^\circ \text{C} \]

Final Answer

  • The molality of the solution is \(\boxed{0.1329 \, \text{m}}\).
  • The boiling point of the solution is \(\boxed{62.19 \, ^\circ \text{C}}\).
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