Calculating Molar Mass Using Vapor Pressure Of A 5% Aqueous Solution

Understanding the colligative properties of solutions is crucial in chemistry, particularly when dealing with non-volatile solutes. This article delves into the concept of vapour pressure lowering and demonstrates how to calculate the molar mass of a non-volatile organic substance dissolved in water. We will use Raoult's Law as a foundation, exploring its implications and practical applications in determining the molar masses of unknown substances. This comprehensive guide aims to provide a step-by-step approach, making the process clear and understandable for students and professionals alike.

Understanding Vapour Pressure and Raoult's Law

Vapour pressure is a fundamental property of liquids that describes the pressure exerted by its vapour when the liquid is in equilibrium with its vapour phase. When a non-volatile solute, such as an organic substance, is dissolved in a volatile solvent like water, the vapour pressure of the solution decreases. This phenomenon is known as vapour pressure lowering, and it's directly related to the concentration of the solute in the solution. Raoult's Law provides a quantitative relationship between the vapour pressure of a solution and the mole fraction of the solvent. Specifically, Raoult's Law states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. Mathematically, this can be represented as:

Psolution = Xsolvent * P0solvent

Where:

• Psolution is the vapour pressure of the solution,
• Xsolvent is the mole fraction of the solvent in the solution, and
• P0solvent is the vapour pressure of the pure solvent.

Raoult's Law is based on the idea that the presence of solute molecules at the surface of the solution reduces the number of solvent molecules that can escape into the vapour phase. The extent of vapour pressure lowering is directly proportional to the number of solute particles present in the solution, regardless of their identity. This colligative property makes vapour pressure lowering a valuable tool for determining the molar masses of unknown substances. By measuring the vapour pressure of a solution and knowing the mass of the solute and solvent, we can calculate the mole fraction of the solvent and, subsequently, the molar mass of the solute. The practical implications of Raoult's Law are vast, ranging from industrial applications in distillation and separation processes to laboratory techniques for characterizing chemical compounds. Understanding and applying this law is essential for any chemist working with solutions and their properties. In the following sections, we will apply Raoult's Law to solve a specific problem and demonstrate the step-by-step process of calculating the molar mass of a non-volatile organic substance.

Problem Statement: Molar Mass Determination

Determining the molar mass of an unknown organic substance dissolved in water is a classic application of colligative properties, particularly vapour pressure lowering. Consider a scenario where we have a 5% aqueous solution of a non-volatile organic substance. This means that 5% of the solution's mass is composed of the organic substance, and the remaining 95% is water. At 373 K (100°C), the solution's vapour pressure is measured to be 745 mmHg. The vapour pressure of pure water at the same temperature is known to be 760 mmHg. Our goal is to calculate the molar mass of the organic substance using this information. This problem highlights the practical application of Raoult's Law in determining the molecular characteristics of a solute. By carefully analyzing the given data and applying the principles of solution chemistry, we can unravel the identity of the unknown substance. The challenge lies in accurately converting the given percentages into mole fractions and then applying Raoult's Law to solve for the molar mass. This problem not only tests our understanding of colligative properties but also our ability to apply mathematical concepts to chemical problems. The solution involves several steps, including calculating the masses of solute and solvent, converting masses to moles, determining mole fractions, and finally, using Raoult's Law to find the molar mass. Each step requires careful attention to detail and a solid understanding of the underlying principles. The following sections will guide you through each of these steps, providing a clear and concise method for solving this type of problem.

Step-by-Step Solution

To solve for the molar mass of the organic substance, we'll break the problem down into manageable steps, ensuring clarity and accuracy in our calculations.

Step 1: Determine the Masses of Solute and Solvent

Firstly, we assume we have 100 g of the solution. Since it's a 5% aqueous solution, this means we have 5 g of the organic substance (solute) and 95 g of water (solvent). This assumption simplifies the calculations by allowing us to work with concrete mass values. The percentage composition directly translates into mass ratios, making it easier to determine the amount of each component in the solution. This step is crucial as it sets the foundation for subsequent calculations involving mole fractions and molar masses. The accuracy of this initial determination directly impacts the final result, so it's important to be precise. Understanding the percentage composition of the solution is key to converting the problem into a form that can be solved using Raoult's Law.

Step 2: Convert Masses to Moles

Next, we need to convert the masses of water and the organic substance into moles. To do this, we use the molar mass of water (approximately 18 g/mol). For water:

• Moles of water = Mass of water / Molar mass of water
• Moles of water = 95 g / 18 g/mol ≈ 5.28 moles

For the organic substance, we don't yet know its molar mass (let's call it M). So, the moles of the organic substance can be expressed as:

• Moles of organic substance = Mass of organic substance / M
• Moles of organic substance = 5 g / M

The conversion of mass to moles is a fundamental step in stoichiometry and is essential for calculating mole fractions, which are required by Raoult's Law. This step bridges the gap between macroscopic measurements (mass) and microscopic quantities (moles), allowing us to work with the number of particles rather than their mass. The molar mass of water is a well-known constant, making this part of the calculation straightforward. However, the molar mass of the organic substance is the unknown we are trying to find, so we express the moles of the organic substance in terms of this unknown variable. This allows us to incorporate it into Raoult's Law and solve for it later.

Step 3: Calculate the Mole Fraction of Water

Now, we calculate the mole fraction of water in the solution. The mole fraction (Xwater) is given by:

• Xwater = Moles of water / (Moles of water + Moles of organic substance)
• Xwater = 5.28 / (5.28 + 5/M)

The mole fraction represents the proportion of water molecules in the solution relative to the total number of molecules (water and organic substance). This dimensionless quantity is crucial for applying Raoult's Law, which relates the vapour pressure of the solution to the mole fraction of the solvent. The calculation involves adding the moles of water and the moles of the organic substance, the latter being expressed in terms of the unknown molar mass M. This step highlights the importance of understanding mole fractions and their role in describing the composition of solutions. The expression for the mole fraction of water will be used in conjunction with Raoult's Law to solve for the molar mass of the organic substance.

Step 4: Apply Raoult's Law

Using Raoult's Law, we have:

• Psolution = Xwater * P0water
• 745 mmHg = [5.28 / (5.28 + 5/M)] * 760 mmHg

Raoult's Law is the cornerstone of this calculation, providing the direct relationship between the vapour pressure of the solution, the mole fraction of the solvent, and the vapour pressure of the pure solvent. By substituting the given values and the expression for the mole fraction of water, we obtain an equation that can be solved for the unknown molar mass M. This step demonstrates the power of Raoult's Law in connecting macroscopic properties (vapour pressure) with microscopic properties (mole fraction and molar mass). The equation obtained is an algebraic expression that needs to be carefully solved to isolate the molar mass, which will be addressed in the next step.

Step 5: Solve for the Molar Mass (M)

Now, we solve the equation for M:

• 745 / 760 = 5.28 / (5.28 + 5/M)
• 0. 9803 = 5.28 / (5.28 + 5/M)
• 0. 9803 * (5.28 + 5/M) = 5.28
• 5. 176 + 4.9015/M = 5.28
• 4. 9015/M = 0.104
• M = 4.9015 / 0.104 ≈ 47.13 g/mol

This step involves algebraic manipulation to isolate the molar mass M. The process includes dividing both sides of the equation, multiplying by the denominator, distributing terms, and finally, solving for M. Each step requires careful attention to ensure accuracy in the calculations. The result, approximately 47.13 g/mol, represents the molar mass of the non-volatile organic substance. This value provides valuable information about the identity of the substance, as it can be compared to known molar masses of organic compounds. This final step demonstrates the culmination of the problem-solving process, where the application of Raoult's Law and algebraic techniques leads to the determination of an unknown molecular property.

Conclusion

In conclusion, the molar mass of the organic substance is approximately 47.13 g/mol. This calculation demonstrates the practical application of Raoult's Law in determining the molar masses of non-volatile solutes. By understanding the colligative properties of solutions, chemists can gain valuable insights into the composition and behavior of mixtures. This method is widely used in various fields, including pharmaceuticals, chemical engineering, and environmental science, for characterizing substances and controlling solution properties. The step-by-step approach outlined in this article provides a clear and concise guide for solving similar problems, emphasizing the importance of accurate measurements and careful calculations. The ability to determine molar masses using colligative properties is a fundamental skill in chemistry, enabling researchers and practitioners to identify and quantify unknown substances. This example showcases the power of Raoult's Law as a tool for understanding and manipulating solutions.

Further Discussion

This discussion extends the principles demonstrated in the previous calculation to broader applications and related concepts in chemistry. The determination of molar mass using vapour pressure lowering is a powerful technique, but it's essential to consider its limitations and the factors that can influence its accuracy. For instance, Raoult's Law is most accurate for ideal solutions, where the interactions between solute and solvent molecules are similar to those between the molecules of each pure component. Deviations from ideality can occur in concentrated solutions or when there are strong interactions between the solute and solvent, such as hydrogen bonding. In such cases, modified versions of Raoult's Law or other colligative properties, such as boiling point elevation or freezing point depression, may provide more accurate results. Furthermore, the experimental determination of vapour pressure requires precise measurements and controlled conditions. Temperature fluctuations, impurities in the solvent, or the presence of volatile solutes can significantly affect the results. Therefore, careful experimental design and data analysis are crucial for obtaining reliable molar mass values. Beyond molar mass determination, the principles of colligative properties are fundamental to understanding various phenomena, such as osmosis, osmotic pressure, and the behavior of biological membranes. These concepts have wide-ranging applications in fields like medicine, food science, and water treatment. Exploring these related topics provides a deeper appreciation for the significance of solution chemistry in both theoretical and practical contexts. The study of solutions and their properties continues to be an active area of research, with ongoing efforts to develop new methods and technologies for characterizing complex mixtures and understanding their behavior under different conditions. This article serves as a starting point for further exploration into the fascinating world of solutions and their colligative properties.