ALEKS - Solving Applied Dilution Problems

Tony St John
13 Feb 202010:17
EducationalLearning
32 Likes 10 Comments

TLDRThis educational video discusses the concept of solving applied dilutions, focusing on two scenarios involving concentration changes due to evaporation. The first example involves calculating the percentage of a lake that has evaporated, given initial and final salt concentrations. The second example addresses the increase in molarity of a solution after solvent evaporation. The instructor emphasizes the importance of understanding the underlying concepts and provides step-by-step calculations to illustrate the process, aiming to clarify this challenging topic.

Takeaways
  • 🧪 The script discusses the concept of solving applied dilutions, which involves understanding the concentration changes in solutions due to evaporation.
  • 📚 The first example involves a geochemist measuring salt concentration in Lake Parsons and nearby lakes to determine the percentage of the lake that has evaporated.
  • 🌊 The initial assumption is that Lake Parsons started with the same salt concentration as the nearby lakes, which is 4.76 grams per liter.
  • 🔍 The script explains that as water evaporates from an isolated lake like Lake Parsons, the concentration of dissolved substances like salt increases.
  • 📉 The calculation to find the percentage of the lake that has evaporated involves comparing the initial and final salt concentrations.
  • 📝 A simplified example is used to illustrate the concept, assuming an initial volume of one liter and calculating the final volume after evaporation.
  • 🔢 The script uses a mathematical approach to solve for the final volume, which represents the remaining water after evaporation.
  • 💧 The percentage of evaporation is then calculated by comparing the initial and final volumes.
  • 🔬 The second example involves a biochemist measuring the molarity of salt in a cell growth medium that has partially evaporated.
  • 🧪 The initial molarity of the solution is 40 millimolar, and after evaporation, the new volume is 8.8 milliliters.
  • 📈 The script demonstrates how to calculate the new concentration of the solution after evaporation, resulting in a higher molarity due to the reduced volume.
Q & A

  • What is the main topic discussed in the video script?

    -The main topic discussed in the video script is solving applied dilutions, particularly focusing on understanding the concepts behind the calculations and how they apply to real-world scenarios involving the concentration of substances in solutions.

  • Why is understanding the concepts behind dilution calculations considered challenging?

    -Understanding the concepts behind dilution calculations is considered challenging because it involves grasping the underlying principles of how concentrations change when a solution evaporates or is otherwise altered, rather than just performing the arithmetic involved in the calculations.

  • What is the scenario presented in the first example of the script?

    -The first scenario involves a geochemist measuring the concentration of salt in Lake Parsons and nearby lakes, and then calculating the percentage of Lake Parsons that has evaporated since it became isolated, based on the change in salt concentration.

  • What was the initial concentration of salt in Lake Parsons compared to the nearby lakes?

    -The initial concentration of salt in Lake Parsons was assumed to be equal to the average concentration in nearby lakes, which was 4.76 grams per liter, before it became isolated.

  • What concentration of salt was measured in Lake Parsons after it became isolated?

    -After Lake Parsons became isolated, the concentration of salt measured was 33.8 grams per liter.

  • How is the evaporation of a solution related to the concentration of its solute?

    -When a solution evaporates, the volume of the solvent decreases while the amount of solute remains the same, leading to an increase in the concentration of the solute as it is now present in a smaller volume of solvent.

  • What method does the script use to calculate the volume of Lake Parsons that has evaporated?

    -The script uses a method similar to cross-multiplication to set up a proportion and solve for the unknown volume that has evaporated, based on the initial and final concentrations of the solute.

  • What was the calculated percentage of Lake Parsons that has evaporated since it became isolated?

    -The calculated percentage of Lake Parsons that has evaporated since it became isolated was 85.9%.

  • What is the second scenario presented in the script?

    -The second scenario involves a biochemist who measures the molarity of salt in a cell growth medium, and then calculates the new concentration after some of the solvent evaporates due to a careless mistake by a graduate student.

  • How does the script calculate the new concentration of the solution after evaporation in the second scenario?

    -The script calculates the new concentration by first determining the amount of salt in the original solution using the initial molarity and volume, and then dividing this amount by the new volume after evaporation to find the new molarity.

  • What was the final concentration of the solution in the second scenario after evaporation?

    -The final concentration of the solution in the second scenario after evaporation was calculated to be 330 millimolar, or 3.3 times 10 to the power of 2 millimolar, when rounded to two significant figures.

Outlines
00:00
🧪 Solving Applied Dilutions: Lake Parsons' Evaporation Mystery

This paragraph introduces the concept of solving applied dilutions with a focus on the challenging nature of understanding the concepts behind the calculations. It presents a geochemistry scenario where the concentration of salt in an isolated lake, Lake Parsons, is compared to nearby lakes. The task is to calculate the percentage of the lake that has evaporated since it became isolated, assuming the initial concentration was equal to the average of nearby lakes. The explanation involves a simplified example using a one-liter volume to demonstrate the concentration process and the calculation of the remaining volume after evaporation, which is found to be 0.141 liters, representing a significant portion of the original volume.

05:00
🌡️ Concentration Shift Due to Evaporation: A Biochemistry Example

The second paragraph delves into a biochemistry context where the molarity of salt in a cell growth medium is measured and then affected by evaporation due to an uncovered container. The original concentration was 40 millimolar in 73 milliliters, and after evaporation, the volume decreased to 8.8 milliliters. The challenge is to determine the new concentration of the solution. By using the original concentration and volume, the total millimoles of salt present are calculated, which is then used to find the new concentration in the reduced volume. The result is a significantly higher concentration of 332 millimolar, demonstrating the effect of solvent reduction on solute concentration. The explanation emphasizes the importance of unit conversion and significant figures in scientific calculations.

10:01
📚 Recap and Reflection on Applied Dilution Problems

The final paragraph serves as a recap and reflection on the process of solving applied dilution problems, emphasizing the importance of understanding the method and the units involved. It suggests that while the problems can be tackled in various ways, the chosen method should clearly demonstrate the process and units. The paragraph concludes by reinforcing the logic behind the concentration increase due to evaporation and encourages the application of the learned concepts to solve similar problems.

Mindmap
Keywords
💡Geochemist
A geochemist is a scientist who studies the chemistry of the Earth and its environment. In the context of the video, the geochemist measures the concentration of salt in different lakes, which is crucial for understanding the evaporation process in Lake Parsons. The script mentions, 'a geochemist measures the concentration of salt dissolved in lake Parsons,' highlighting the role of geochemists in environmental analysis.
💡Concentration
Concentration in this video refers to the amount of a substance (in this case, salt) dissolved in a given volume of a solvent (water). It is measured in grams per liter. The script discusses how the concentration of salt in Lake Parsons is 33.8 grams per liter, contrasting with the concentration in nearby lakes, which is 4.76 grams per liter. This concept is central to understanding the dilution and evaporation processes.
💡Isolated Lake
An isolated lake is a body of water that is not connected to other water bodies. The script states that Lake Parsons is an isolated lake, implying that it has no inflow or outflow of water. This isolation is significant because it means that the evaporation of water in the lake will lead to an increase in the concentration of dissolved substances like salt.
💡Evaporation
Evaporation is the process by which water changes from a liquid to a gas or vapor and enters the atmosphere. In the video, the script describes how the evaporation of water in Lake Parsons has led to an increase in the concentration of salt. The concept is illustrated by the script's mention of 'a bunch of the liquid evaporated' and the subsequent increase in salt concentration.
💡Salt Concentration
Salt concentration is the amount of salt present in a given volume of water. The video discusses how the salt concentration in Lake Parsons is initially the same as in nearby lakes but changes over time due to evaporation. The script uses the term to explain the process of concentration increase, stating 'the concentration of salt in Lake Parsons both before it became isolated was equal to the average salt concentration in nearby non-isolated Lakes.'
💡Cross Multiplication
Cross multiplication is a mathematical technique used to solve proportions. In the video, the script uses this method to calculate the volume of water that would result in a specific salt concentration. The script explains, 'this is sort of like cross multiplication you know, solve for X situation,' demonstrating how to use this technique to find the volume that would lead to the observed concentration in Lake Parsons.
💡Molarity
Molarity is a measure of concentration that indicates the number of moles of solute per liter of solution. The script introduces a scenario where a biochemist measures the molarity of salt in a cell growth medium to be 40 millimolar. This term is crucial for understanding the changes in concentration due to solvent evaporation.
💡Solvent
In the context of the video, the solvent refers to the water in which the salt is dissolved. The script discusses how the evaporation of the solvent (water) leads to an increase in the concentration of the solute (salt). The term is used to explain the process of concentration change, as in 'we've evaporated some of that solvent so we've concentrated our solution here.'
💡Millimolar
Millimolar is a unit of concentration equal to one-thousandth of a molar (mmol/L). The script uses this term to describe the concentration of salt in the cell growth medium, stating it was initially 40 millimolar. This term is important for understanding the initial and final concentrations in the evaporation example.
💡Significant Figures
Significant figures are the digits in a number that carry meaning concerning the precision of that number. The script mentions the importance of reporting the concentration with the correct number of significant figures, suggesting that '3.3 times 10 to the 2nd millimolar' is a better answer than '330 millimolar' due to the precision of the measurement.
Highlights

Introduction to the topic of solving applied dilutions with a focus on challenging problem-solving concepts.

Explanation of the straightforward nature of calculations in applied dilutions, contrasting with the complexity of understanding the underlying concepts.

Presentation of two different instances of applied dilution problems to demonstrate varying levels of complexity.

Description of a geochemist measuring salt concentration in Lake Parsons and nearby lakes to understand evaporation effects.

Assumption that the initial salt concentration in Lake Parsons was equal to the average of nearby lakes before isolation.

Calculation of the percentage of Lake Parsons that has evaporated since becoming isolated, using a simplified example.

Illustration of the concept of concentration changes due to evaporation, using the analogy of a liter of solution.

Use of a cross-multiplication method to solve for the new volume after evaporation, leading to a concentrated salt solution.

Explanation of the significance of the calculated volume representing the remaining liquid after evaporation.

Conversion of the calculated volume into a percentage to determine the amount of Lake Parsons that has evaporated.

Introduction of a second example involving a biochemist measuring molarity in a cell growth medium.

Scenario of a careless graduate student causing solvent evaporation from an uncovered container, leading to concentration changes.

Methodology for determining the new concentration of the solution after evaporation, using the original molarity and volume.

Calculation of the original amount of salt in the solution based on the initial molarity and volume.

Determination of the new concentration after evaporation by dividing the original amount of salt by the new volume.

Emphasis on the importance of checking that the new concentration is higher than the original, confirming the concentration process.

Discussion on the significance of understanding the principles behind dilution and concentration in practical applications.

Conclusion summarizing the importance of visual aids and clear problem-solving steps in grasping complex dilution concepts.

Transcripts

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Dilution ProblemsGeochemistryConcentrationEvaporationSolution AnalysisProblem SolvingEducational ContentScientific MethodMolarityChemistry Tutorial