Problem 96 - General Chemistry: Principles and Modern Applications

Question

Nitrobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2},\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) are completely miscible in each other. Other properties of the two liquids are nitrobenzene: \(\mathrm{fp}=5.7^{\circ} \mathrm{C}, K_{\mathrm{f}}=\) \(8.1^{\circ} \mathrm{C} m^{-1} ;\) benzene: \(\mathrm{fp}=5.5^{\circ} \mathrm{C}, K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} m^{-1} . \mathrm{It}\) is possible to prepare two different solutions with these two liquids having a freezing point of \(0.0^{\circ} \mathrm{C}\) What are the compositions of these two solutions, expressed as mass percent nitrobenzene?

Step-by-Step Solution

Answer

The two solutions with a freezing point of \(0.0^{\circ}C\) are composed, by mass, of approximately 71% nitrobenzene and 29% benzene for the nitrobenzene solution, and 92% nitrobenzene and 8% benzene for the benzene solution.

Step 1 - Calculate the depression of freezing point for both substances

Since the freezing point of both solutions is $0.0^{\circ}C$, the depression of freezing point \(\Delta T_{f}\) for nitrobenzene is \(5.7^{\circ}C - 0.0^{\circ}C = 5.7^{\circ}C\) and for benzene is \(5.5^{\circ}C - 0.0^{\circ}C = 5.5^{\circ}C.\)

Step 2 - Calculate the molality of each solution

Using the freezing point depression formula: \(\Delta T_{f} = K_{f} * m\), where \(m\) is the molality, \(K_{f}\) is the freezing point depression constant and \(\Delta T_{f}\) is the change in freezing point. Solving for \(m\): For nitrobenzene: \(m = \Delta T_{f} / K_{f} = 5.7 ^{\circ}C / 8.1^{\circ}C * m^{-1} = 0.7037 m\); For benzene: \(m = \Delta T_{f} / K_{f} = 5.5 ^{\circ}C / 5.12^{\circ}C * m^{-1} = 1.0742 m\)

Step 3 - Calculate the mass percent Nitrobenzene

First calculate the moles of nitrobenzene in each solution. Use the molecular weight of nitrobenzene, 123.11 g/mol: For nitrobenzene solution: \(\text{Moles of nitrobenzene} = \text{Molality} * \text{Mass of benzene} = 0.7037 m * 100 g = 70.37 \text{mol}\); For benzene solution: \(\text{Moles of nitrobenzene} = \text{Molality} * \text{Mass of benzene} = 1.0742 m * 100 g = 107.42 \text{mol}\). Now calculate the mass of nitrobenzene in each solution: For nitrobenzene solution: \(\text{Mass of nitrobenzene} = \text{Moles of nitrobenzene} * \text{Molar mass of nitrobenzene}\) ; For benzene solution: \(\text{Mass of nitrobenzene} = \text{Moles of nitrobenzene} * \text{Molar mass of nitrobenzene}\). Finally, calculate the mass percent of nitrobenzene in each solution: Mass percent = (\(\text{Mass of nitrobenzene} / (\text{Mass of nitrobenzene} + \text{Mass of benzene})\) )* 100

Key Concepts

Molality Calculation

When we talk about molality, we're looking at the concentration of a solution. It's expressed as the number of moles of solute per kilogram of solvent. This is a super handy way to measure concentration because it doesn't change with temperature. Here's how we can calculate it:
\[ m = \frac{\Delta T_{f}}{K_{f}} \]where \( \Delta T_{f} \) is the depression in freezing point, and \( K_{f} \) is the freezing point depression constant.
For nitrobenzene, using the values given, the molality \( m \) is calculated as:
\[ m = \frac{5.7 ^{\circ}C}{8.1^{\circ}C \, m^{-1}} = 0.7037 \, m \]For benzene, the calculation goes:
\[ m = \frac{5.5 ^{\circ}C}{5.12^{\circ}C \, m^{-1}} = 1.0742 \, m \]This helps us know how concentrated nitrobenzene is in each solution.

Mass Percent Composition

Mass percent tells you how much of one substance is in a mixture compared to the total weight of the mixture. It's like a pie chart showing the exact share of each ingredient. To find mass percent:
1. Calculate the mass of the solute (here, nitrobenzene) using molality and the solvent's mass, which is benzene in this case.2. Determine the total mass of the solution.3. Use the formula:

• Mass percent = \(\frac{\text{Mass of nitrobenzene}}{\text{Mass of nitrobenzene} + \text{Mass of benzene}} * 100\)

This tells you the percentage of the entire solution that is made up of nitrobenzene. Understanding mass percent helps in grasping the solution's concentration compared to others.

Miscibility of Liquids

Miscibility refers to how well two liquids mix, forming a uniform solution. Nitrobenzene and benzene are known to be completely miscible; they blend perfectly into one another at any proportion. This is crucial for solutions where the ingredients need to be incorporated seamlessly without any separation, like how sugar dissolves into tea.
Miscibility is vital when creating solutions because if two liquids are not miscible, you can't form a homogeneous mixture, and the purpose of creating that solution might be defeated. In everyday life, mixtures like oil and water show immiscibility—they don’t mix! By understanding which liquids are miscible, you can better predict the behavior of mixtures and solutions in both experimental and real-world scenarios.