Calculating Molality A Step-by-Step Guide For 10% NaOH Solution
Hey everyone! Today, we're diving into a fascinating chemistry problem involving molality calculations. We'll be working with a 10% by mass NaOH solution that has a density of 1.06 g/mL. Our main goal is to figure out the molality of this solution. Molality, as you might remember, is a measure of the concentration of a solution expressed in moles of solute per kilogram of solvent. So, let's break down this problem step by step and make sure we understand each part of the calculation.
Understanding the Problem: What Do We Know?
First off, let's highlight what we already know. We're dealing with a 10% by mass NaOH solution. This means that for every 100 grams of the solution, 10 grams are NaOH (our solute), and the remaining 90 grams are water (our solvent). We also know that the density of the solution is 1.06 g/mL. This is super important because it allows us to convert between volume and mass. To nail this calculation, we'll need to convert grams of NaOH to moles and grams of water to kilograms. Remember, molality is expressed in moles per kilogram, so units are key here!
Step 1: Choosing a Convenient Volume
To make our calculations easier, let's start with a convenient volume of the solution. A good choice is 100 mL. Why 100 mL? Because it simplifies the math! Using the density, we can easily calculate the mass of this volume of solution. We know that Density = Mass / Volume, so Mass = Density × Volume. In our case, the mass of 100 mL of the solution is 1.06 g/mL × 100 mL = 106 grams. This gives us a starting point to work with.
Step 2: Calculating the Mass of NaOH and Water
Now that we know the total mass of the solution, we can figure out the mass of NaOH and water. Since it's a 10% by mass solution, 10% of the 106 grams is NaOH. So, the mass of NaOH = 0.10 × 106 grams = 10.6 grams. The rest of the solution is water, so the mass of water = 106 grams (total) - 10.6 grams (NaOH) = 95.4 grams. These values are crucial for the next steps.
Step 3: Converting Grams of NaOH to Moles
Next, we need to convert the mass of NaOH to moles because molality is defined in terms of moles of solute. To do this, we'll use the molar mass of NaOH. The molar mass of NaOH is approximately 40 g/mol (23 g/mol for Na + 16 g/mol for O + 1 g/mol for H). So, the number of moles of NaOH = mass of NaOH / molar mass of NaOH = 10.6 grams / 40 g/mol = 0.265 moles. Now we know how much NaOH we're working with in terms of moles.
Step 4: Converting Grams of Water to Kilograms
The final conversion we need to make is from grams of water to kilograms because molality uses kilograms of solvent. We know we have 95.4 grams of water. To convert this to kilograms, we divide by 1000 (since there are 1000 grams in a kilogram). So, the mass of water in kilograms = 95.4 grams / 1000 = 0.0954 kg. We're almost there, guys!
Step 5: Calculating the Molality
Now for the grand finale – calculating the molality! Molality is defined as moles of solute per kilogram of solvent. We have 0.265 moles of NaOH and 0.0954 kg of water. So, molality = moles of NaOH / kilograms of water = 0.265 moles / 0.0954 kg = 2.78 m (approximately). And there you have it! The molality of the 10% by mass NaOH solution is approximately 2.78 m.
Key Concepts and Formulas
Let's quickly recap the key concepts and formulas we used:
- Molality: Moles of solute / kilograms of solvent
- Density: Mass / Volume
- Molar Mass: The mass of one mole of a substance (found by adding up the atomic masses of all the atoms in the molecule)
- Percentage by Mass: (Mass of solute / Mass of solution) × 100
Understanding these concepts is super important for tackling any molality problem. Make sure you're comfortable with these formulas and how to apply them.
Common Mistakes to Avoid
When calculating molality, there are a few common mistakes that students often make. Let's go over them so you can avoid these pitfalls:
- Mixing Up Molality and Molarity: This is a big one! Molality is moles per kilogram of solvent, while molarity is moles per liter of solution. They sound similar, but they're different. Always double-check which one you're asked to calculate.
- Forgetting to Convert Units: Remember, molality is in moles per kilogram. Don't forget to convert grams to kilograms and use the correct molar mass to convert grams to moles.
- Using the Wrong Mass for the Solvent: Make sure you're using the mass of the solvent (in our case, water) and not the mass of the entire solution. This is a common mistake that can throw off your calculations.
- Incorrectly Calculating Molar Mass: Always double-check your molar mass calculations. A small error here can lead to a big mistake in your final answer.
Real-World Applications of Molality
So, why is molality important in the real world? Well, molality is particularly useful because it doesn't change with temperature. Unlike molarity, which is volume-dependent and can change with temperature fluctuations, molality remains constant. This makes it super useful in situations where temperature varies, such as in experiments or industrial processes. Think about preparing solutions that need to maintain consistent concentrations under different conditions – molality is your go-to measure!
Applications in Chemistry and Beyond
- Cryoscopy and Ebullioscopy: Molality is used in colligative properties calculations, such as freezing point depression (cryoscopy) and boiling point elevation (ebullioscopy). These properties depend on the number of solute particles in a solution, and molality provides an accurate measure of this.
- Pharmaceuticals: In the pharmaceutical industry, precise concentrations are crucial. Molality is often used to ensure the accuracy of drug formulations, especially when temperature stability is a concern.
- Chemical Research: Researchers use molality when preparing solutions for experiments where temperature might fluctuate, ensuring consistent results.
- Industrial Processes: In various industrial processes, molality helps maintain the consistency of solutions, which is vital for the quality and efficiency of the processes.
Practice Problems
To really nail this concept, let's try a few practice problems. Remember, practice makes perfect!
- What is the molality of a solution prepared by dissolving 25 grams of glucose (C6H12O6) in 500 grams of water?
- A solution of sulfuric acid (H2SO4) has a density of 1.84 g/mL and is 98% by mass. What is the molality of this solution?
- If you dissolve 15 grams of NaCl in 150 grams of water, what is the molality of the solution?
Work through these problems, and don't hesitate to refer back to the steps we discussed earlier. The more you practice, the more confident you'll become with molality calculations.
Conclusion
Calculating molality might seem tricky at first, but with a step-by-step approach and a good understanding of the key concepts, it becomes much easier. Remember to focus on the definitions, pay attention to units, and practice, practice, practice! By breaking down the problem into smaller, manageable steps, you can solve even the most challenging molality calculations. Keep up the great work, and you'll be a molality master in no time! So, guys, keep practicing, and you'll ace those chemistry problems!
Understanding molality is crucial in various fields, from chemistry labs to industrial applications. Its temperature independence makes it a reliable measure for consistent solution concentrations, ensuring accuracy in experiments and processes alike.
In summary, we've covered the step-by-step process of calculating molality for a 10% NaOH solution, highlighted key formulas and concepts, discussed common mistakes to avoid, explored real-world applications, and provided practice problems to reinforce your understanding. Keep these strategies in mind, and you'll be well-equipped to tackle any molality problem that comes your way.
