Calculating Vapor Pressure: Sucrose And Glucose Solutions

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Hey guys! Let's dive into a classic chemistry problem: calculating the vapor pressure of solutions. We'll tackle two examples, one involving sucrose and the other glucose. This will give you a solid understanding of how to apply Raoult's Law and how to approach these types of calculations. So, grab your calculators and let's get started! We'll break down each step, making it super easy to follow. Whether you're a student, a science enthusiast, or just curious, this guide will help you master the concepts. Ready? Let's go!

Problem 1: Vapor Pressure of a Sucrose Solution

Determining the vapor pressure of a sucrose solution is a fundamental application of Raoult's Law. This law is super useful for understanding how the presence of a solute (like sucrose) affects the vapor pressure of a solvent (like water). The specific problem we are going to solve is: What is the vapor pressure, in mmHg, of a 4-gram sucrose solution at 25°C, given that the vapor pressure of pure water at that temperature is 34.80 mmHg, and the Mr (molecular weight) of water is 18 and sucrose is 342?

Okay, so the first thing we need to understand is Raoult's Law. This law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. In simpler terms, the more solvent you have relative to the solute, the closer the solution's vapor pressure will be to that of the pure solvent. In this case, we're using water as our solvent. To solve this problem, we're going to follow a few steps.

First, we'll calculate the moles of sucrose. Then, we'll calculate the moles of water. After that, we'll determine the mole fraction of water in the solution. Finally, we'll use Raoult's Law to calculate the vapor pressure of the solution. It might sound like a lot, but trust me, it's quite straightforward when broken down. Let's start with calculating the moles of sucrose. We know that we have 4 grams of sucrose, and we know the Mr of sucrose is 342. To calculate the moles of sucrose, we'll use the formula: moles = mass / Mr. So, moles of sucrose = 4 g / 342 g/mol = 0.0117 moles. Now we know how many moles of sucrose we have in the solution. This gives us a good start for the problem.

Next, we need to figure out the moles of water. This might seem tricky because we only know the mass of sucrose (4 grams) and the total volume isn't given. However, we can approximate the mass of the solution by assuming that the mass of the solution is equal to the mass of sucrose plus the mass of water. It is safe to say that 4 grams of sucrose is negligible compared to the amount of water, so we can assume the solution is mostly water. To determine the moles of water, we'll start by assuming we have a certain amount of water in grams. We'll assume the amount of the solution in grams is approximately equal to the grams of water, since we only know the grams of sucrose which is 4 grams. We'll use the density of water, which is approximately 1 g/mL, and convert this to mass. Now that we have an estimate for the mass of the water, we can calculate the moles of water. The Mr of water is 18 g/mol. If we assume we have 100 g of water (just for simplicity – the exact mass isn't critical, but it makes the calculations easier), then moles of water = 100 g / 18 g/mol = 5.56 moles. This is the second most important step in solving the problem.

With the moles of sucrose and water calculated, we can now compute the mole fraction of water. The mole fraction (X) is the ratio of the moles of the solvent to the total moles of solute and solvent. In other words, it tells us the proportion of the solvent in the solution, expressed as a fraction. To calculate the mole fraction of water (Xwater), we'll use the formula: Xwater = moles of water / (moles of water + moles of sucrose). So, Xwater = 5.56 moles / (5.56 moles + 0.0117 moles) = 0.998. The mole fraction of the water is quite close to 1, which makes sense because sucrose is a minor component. This step is very important in the problem to allow us to calculate the vapor pressure.

Finally, it's time to apply Raoult's Law to find the vapor pressure of the sucrose solution. Raoult's Law states that the vapor pressure of the solution (P) is equal to the mole fraction of the solvent (Xwater) multiplied by the vapor pressure of the pure solvent (P°water). The formula looks like this: P = Xwater * P°water. We know Xwater is 0.998, and the vapor pressure of pure water (P°water) at 25°C is given as 34.80 mmHg. Now we can calculate the vapor pressure of the sucrose solution: P = 0.998 * 34.80 mmHg = 34.73 mmHg. So, the vapor pressure of the sucrose solution at 25°C is approximately 34.73 mmHg. Done! You have just solved the problem.

Problem 2: Vapor Pressure of a Glucose Solution

Calculating the vapor pressure of a glucose solution is a good example of another application of Raoult's Law. Let's solve the problem: Calculate the vapor pressure of a 72% glucose solution in water at 25°C. We know the vapor pressure of pure water at 25°C is 34.80 mmHg. In this problem, we're given the percentage of glucose in the solution by mass. This means that 72% of the mass is glucose, and the remaining 28% is water. The steps we need to follow here are very similar to the previous problem, but we'll start with the percentage composition. We know the vapor pressure of pure water at 25°C. We'll assume a 100 g solution as our base.

First, we'll determine the mass of glucose and water based on the percentage composition. Then, we'll calculate the moles of glucose and water. After that, we'll find the mole fraction of water in the solution. Finally, we'll apply Raoult's Law to calculate the vapor pressure of the solution. It's going to be easy, trust me. Here's how we'll do it. As mentioned, we will assume we have 100 g of the solution. Since the solution is 72% glucose, we have 72 g of glucose. This means that we have 28 g of water (100 g - 72 g). We are going to take it step by step, so don't worry.

Next, we'll calculate the moles of glucose and water. The Mr of glucose is 180 g/mol (you can find this on the periodic table). The Mr of water is still 18 g/mol. To calculate the moles of glucose, we'll use the formula: moles = mass / Mr. So, moles of glucose = 72 g / 180 g/mol = 0.40 moles. Then, to calculate the moles of water, we use the same formula: moles of water = 28 g / 18 g/mol = 1.56 moles. This is a very important step, since you'll need these values for the next step. Now we know how many moles of glucose and water we have in the solution. Now we can start with the mole fraction.

Now, we need to find the mole fraction of water in the solution. Remember, the mole fraction is the ratio of the moles of water to the total moles of solute and solvent. The mole fraction of water is the most important factor to determine the vapor pressure of the solution. To calculate the mole fraction of water (Xwater), we'll use the formula: Xwater = moles of water / (moles of water + moles of glucose). So, Xwater = 1.56 moles / (1.56 moles + 0.40 moles) = 0.796. This tells us that 79.6% of the molecules in the solution are water. Good, we now have the main factors to solve this problem.

Finally, we can use Raoult's Law to calculate the vapor pressure of the glucose solution. As we did before, Raoult's Law states that the vapor pressure of the solution (P) is equal to the mole fraction of the solvent (Xwater) multiplied by the vapor pressure of the pure solvent (P°water). The formula is: P = Xwater * P°water. We know Xwater is 0.796, and the vapor pressure of pure water (P°water) at 25°C is 34.80 mmHg. Now we can calculate the vapor pressure of the glucose solution: P = 0.796 * 34.80 mmHg = 27.71 mmHg. So, the vapor pressure of the 72% glucose solution at 25°C is approximately 27.71 mmHg. And that's it!

Conclusion

So, there you have it, guys! We've successfully calculated the vapor pressure of a sucrose solution and a glucose solution. By understanding Raoult's Law and following these step-by-step calculations, you can solve similar problems with confidence. Remember that the key is to break down the problem into smaller, manageable steps, and always pay attention to the units. Keep practicing, and you'll become a pro in no time! If you have any other chemistry problems you want to solve, don't hesitate to ask! Keep learning and have fun with chemistry! You got this!