Freezing Point Depression Calculation For Pyrazine In Carbon Tetrachloride Solution

In the realm of chemistry, understanding the behavior of solutions is paramount. One fascinating aspect is the freezing point depression, a colligative property that describes the lowering of a solvent's freezing point upon the addition of a solute. This phenomenon has practical applications, from antifreeze in car radiators to de-icing salts on roads. In this comprehensive article, we will delve into the process of calculating the freezing point depression of a solution, using a specific example involving carbon tetrachloride ($CCl_4$) and pyrazine $\left( C_4 H_4 N_2 \right)$. We will break down the concepts, formulas, and steps involved, providing a clear and concise guide for students, researchers, and anyone interested in this fundamental chemical principle.

Freezing point depression is a colligative property, meaning it depends on the number of solute particles present in a solution, rather than the nature of those particles. When a solute is dissolved in a solvent, it disrupts the solvent's crystal lattice structure, making it more difficult for the solvent to solidify. Consequently, the freezing point of the solution is lower than that of the pure solvent. This depression is directly proportional to the molality of the solute in the solution.

The key equation governing freezing point depression is:

$$\Delta T_f = K_f \cdot m$$

Where:

• \Delta T_f$ is the freezing point depression, the difference between the freezing point of the pure solvent and the freezing point of the solution (in °C).

• K_f$ is the cryoscopic constant or freezing point depression constant, a characteristic property of the solvent (in °C·kg/mol).

• m$ is the molality of the solution, defined as the number of moles of solute per kilogram of solvent (in mol/kg).

To effectively apply this equation, we must first understand the meaning and significance of each component. The freezing point depression ($\Delta T_f$) is the value we ultimately seek to determine. The cryoscopic constant ($K_f$) is a solvent-specific value, often found in reference tables. Molality (m) requires us to calculate the moles of solute and the mass of the solvent in kilograms.

Let's consider the following problem:

The freezing point of $CCl_4$ (carbon tetrachloride) is $-22.92^{\circ}C$. Calculate the freezing point of the solution prepared by dissolving 17.3 g of pyrazine $\left( C_4 H_4 N_2 \right)$ in 1,160 g of $CCl_4$. The $K_f$ for $CCl_4$ is 29.8 $^{\circ}C \cdot kg/mol$.

This problem provides us with all the necessary information to calculate the freezing point of the solution. We know the freezing point of the pure solvent ($CCl_4$), the mass of the solute (pyrazine), the mass of the solvent ($CCl_4$), and the cryoscopic constant for $CCl_4$. Our task is to utilize this information to determine the freezing point depression and, subsequently, the freezing point of the solution.

To solve this problem, we will follow these steps:

1. Calculate the moles of solute (pyrazine).
2. Calculate the molality of the solution.
3. Calculate the freezing point depression ($\Delta T_f$).
4. Calculate the freezing point of the solution.

Step 1: Calculating Moles of Solute (Pyrazine)

To determine the molality of the solution, we first need to calculate the number of moles of pyrazine dissolved in $CCl_4$. To do this, we will use the formula:

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$

The mass of pyrazine is given as 17.3 g. We need to calculate the molar mass of pyrazine ($C_4 H_4 N_2$). The molar mass is the sum of the atomic masses of all the atoms in the molecule. The atomic masses are:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.01 g/mol
• Nitrogen (N): 14.01 g/mol

Therefore, the molar mass of pyrazine is:

$$(4 \times 12.01) + (4 \times 1.01) + (2 \times 14.01) = 48.04 + 4.04 + 28.02 = 80.10 \text{ g/mol}$$

Now we can calculate the moles of pyrazine:

$$\text{Moles of pyrazine} = \frac{17.3 \text{ g}}{80.10 \text{ g/mol}} = 0.216 \text{ mol}$$

Step 2: Calculating Molality of the Solution

Next, we calculate the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{\text{Moles of solute}}{\text{Kilograms of solvent}}$$

We have already calculated the moles of pyrazine (0.216 mol). The mass of the solvent ($CCl_4$) is given as 1,160 g. We need to convert this to kilograms:

$$1,160 \text{ g} = 1.160 \text{ kg}$$

Now we can calculate the molality:

$$m = \frac{0.216 \text{ mol}}{1.160 \text{ kg}} = 0.186 \text{ mol/kg}$$

Step 3: Calculating Freezing Point Depression

Now that we have the molality of the solution, we can calculate the freezing point depression using the formula:

$$\Delta T_f = K_f \cdot m$$

We are given the cryoscopic constant ($K_f$) for $CCl_4$ as 29.8 °C·kg/mol. We have calculated the molality (m) as 0.186 mol/kg. Plugging these values into the formula:

$$\Delta T_f = 29.8 \frac{{}^{\circ}C \cdot \text{kg}}{\text{mol}} \times 0.186 \frac{\text{mol}}{\text{kg}} = 5.54 {}^{\circ}C$$

This result tells us that the freezing point of the solution will be lowered by 5.54 °C compared to the pure solvent.

Step 4: Calculating the Freezing Point of the Solution

Finally, we can calculate the freezing point of the solution. The freezing point depression ($\Delta T_f$) is the difference between the freezing point of the pure solvent and the freezing point of the solution:

$$\Delta T_f = T_{f, \text{pure solvent}} - T_{f, \text{solution}}$$

We are given the freezing point of pure $CCl_4$ as -22.92 °C. We have calculated the freezing point depression as 5.54 °C. We can rearrange the formula to solve for the freezing point of the solution:

$$T_{f, \text{solution}} = T_{f, \text{pure solvent}} - \Delta T_f$$

$$T_{f, \text{solution}} = -22.92 {}^{\circ}C - 5.54 {}^{\circ}C = -28.46 {}^{\circ}C$$

Therefore, the freezing point of the solution prepared by dissolving 17.3 g of pyrazine in 1,160 g of $CCl_4$ is -28.46 °C.

In this article, we have explored the concept of freezing point depression and demonstrated a step-by-step calculation to determine the freezing point of a solution. By understanding the principles of colligative properties and applying the relevant formulas, we can accurately predict the behavior of solutions. This knowledge is crucial in various fields, including chemistry, materials science, and engineering.

We started by defining freezing point depression and its relationship to molality and the cryoscopic constant. We then meticulously worked through an example problem, calculating the moles of solute, molality of the solution, freezing point depression, and finally, the freezing point of the solution. This process highlights the importance of careful unit conversions and accurate application of formulas.

Understanding freezing point depression allows us to tailor solutions for specific applications. For instance, antifreeze in car radiators utilizes the principle of freezing point depression to prevent the coolant from freezing in cold temperatures. Similarly, salts are used to de-ice roads by lowering the freezing point of water.

By mastering the concepts and calculations presented in this article, you will gain a solid foundation for understanding the behavior of solutions and their properties. The ability to calculate freezing point depression is a valuable skill for anyone working in the chemical sciences or related fields. Continue to practice and explore further applications of these principles to deepen your understanding and broaden your expertise.

To further enhance your understanding of colligative properties and freezing point depression, consider exploring these additional topics:

• Boiling Point Elevation: Another colligative property that describes the increase in a solvent's boiling point upon the addition of a solute.
• Osmotic Pressure: A colligative property related to the movement of solvent across a semipermeable membrane.
• Raoult's Law: A law that describes the vapor pressure of a solution in relation to the vapor pressure of the pure solvent and the mole fraction of the solute.
• Applications of Colligative Properties: Explore real-world applications of colligative properties in various industries and technologies.

By delving into these related topics, you will gain a more comprehensive understanding of solutions and their unique behaviors. Understanding these concepts will not only help in academic pursuits but also in practical applications, such as designing experiments and interpreting results in various scientific fields. The more you explore these topics, the better equipped you will be to tackle complex problems and contribute to advancements in science and technology.

To solidify your understanding of freezing point depression, try solving the following practice problems:

1. What is the freezing point of a solution prepared by dissolving 10.0 g of glucose ($C_6H_{12}O_6$) in 100.0 g of water? ($K_f$ for water is 1.86 °C·kg/mol).
2. A solution containing 2.50 g of an unknown non-electrolyte solute in 50.0 g of benzene freezes at 4.3 °C. The freezing point of pure benzene is 5.5 °C, and its $K_f$ is 5.12 °C·kg/mol. What is the molar mass of the solute?
3. How many grams of NaCl must be added to 1.00 kg of water to lower the freezing point by 2.0 °C? ($K_f$ for water is 1.86 °C·kg/mol, and assume complete dissociation of NaCl).

Working through these problems will provide you with hands-on experience in applying the concepts and formulas discussed in this article. Remember to break down each problem into steps, identify the given information, and select the appropriate equations to solve for the unknowns. Practice makes perfect, and the more you practice, the more confident you will become in your ability to tackle freezing point depression problems.

By mastering the principles of freezing point depression, you are not only enhancing your understanding of chemistry but also gaining valuable problem-solving skills that can be applied to a wide range of scientific and engineering challenges. Keep exploring, keep practicing, and keep pushing the boundaries of your knowledge.