How to Calculate Molality of Solutions Examples, Practice Problems, Equation, Shortcut, Explanation

Conquer Chemistry
21 Jun 202005:25
EducationalLearning
32 Likes 10 Comments

TLDRThis educational video explains the concept of molality, a temperature-independent unit of concentration, contrasting it with molarity. The presenter uses two examples to demonstrate how to calculate molality: the first involves dissolving 128 grams of KBr in 925 grams of water, while the second is more complex, involving 15 grams of H2SO4 in 575 milliliters of water at 80°C, requiring the use of water's density at that temperature. The video aims to clarify the process of converting grams to moles and milliliters to kilograms before applying the molality formula. The host also offers a free guide with chemistry study tips on their website.

Takeaways
  • 🔍 Molality is a unit of concentration similar to molarity but is temperature independent.
  • 📘 The formula for molality is the moles of solute divided by the kilograms of solvent.
  • 🧪 Example 1: To find the molality of a KBr solution, convert grams of KBr to moles and grams of water to kilograms, then apply the formula.
  • 📚 KBr's molar mass is approximately 119 g/mol, which is used to convert grams to moles in the first example.
  • 📉 Convert grams of water to kilograms by dividing by 1000, as there are 1000 grams in a kilogram.
  • 🧪 Example 2: For a sulfuric acid (H2SO4) solution, identify the solute and solvent, then convert grams to moles and milliliters to kilograms.
  • 🔢 The molar mass of H2SO4 is approximately 98.07 g/mol, used to calculate moles from grams in the second example.
  • 💧 Use the density of water at 80 degrees Celsius (0.997 g/mL) to convert volume to mass in the second example.
  • ⚖️ Convert the mass of water from grams to kilograms by dividing by 1000 before using it in the molality equation.
  • 📝 Keep track of significant figures, as demonstrated by rounding the molality of the H2SO4 solution to two significant figures.
  • 📚 For additional chemistry study tips and tricks, a free guide called 'Twelve Secrets to Acing Chemistry' is available on the speaker's website.
Q & A
  • What is molality and how does it differ from molarity?

    -Molality is a unit of concentration that measures the amount of solute in a solution, similar to molarity. The key difference is that molality is expressed in moles of solute per kilogram of solvent, whereas molarity is expressed in moles of solute per liter of solution. Unlike molarity, molality is temperature-independent and does not change when the temperature is altered.

  • What is the formula to calculate molality?

    -The formula to calculate molality is given by the lowercase 'm', which is the moles of the solute (M) divided by the kilograms of solvent (kg).

  • In the first example, what is the solute and solvent in the KBR solution?

    -In the first example, KBR is the solute and water is the solvent. The solute is the substance present in a lesser amount, while the solvent is the one in the greater amount.

  • How do you convert grams to moles for the solute in the first example?

    -To convert grams to moles for the solute, you divide the mass of the solute (in grams) by its molar mass. In the case of KBR, the molar mass is approximately 119 g/mol. So, 128 grams of KBR divided by 119 g/mol gives approximately 1.0708 moles.

  • How do you convert grams of water to kilograms in the first example?

    -To convert grams to kilograms, you divide the mass of the water (in grams) by 1000, since there are 1000 grams in a kilogram. In the example, 925 grams of water divided by 1000 gives 0.925 kilograms.

  • What is the calculated molality of the KBR solution in the first example?

    -The calculated molality of the KBR solution is approximately 1.08 moles divided by 0.925 kilograms, which equals approximately 1.16 mol/kg.

  • In the second example, what is the solute and solvent in the H2SO4 solution?

    -In the second example, H2SO4 is the solute and water is the solvent. The solute is the substance present in a lesser amount, while the solvent is the one in the greater amount.

  • How do you calculate the moles of H2SO4 in the second example?

    -To calculate the moles of H2SO4, you take the mass of H2SO4 (15 grams) and divide it by its molar mass (approximately 98.07 g/mol). This gives approximately 0.153 moles of H2SO4.

  • How is the mass of water in kilograms calculated in the second example using density?

    -The mass of water is calculated by multiplying the volume of water (575 mL) by the density of water at 80 degrees Celsius (0.997 g/mL). Then, convert grams to kilograms by dividing by 1000 (since there are 1000 grams in a kilogram), which gives approximately 0.573 kilograms.

  • What is the calculated molality of the H2SO4 solution in the second example?

    -The calculated molality of the H2SO4 solution is approximately 0.153 moles divided by 0.573 kilograms, which equals approximately 0.27 mol/kg.

  • What is the significance of using density in the second example?

    -The significance of using density in the second example is to account for the fact that the mass of water changes with temperature. By using the density at 80 degrees Celsius, you can accurately convert the volume of water to its mass in kilograms, which is necessary for calculating molality.

  • Where can I find more tips and tricks to improve my chemistry studies?

    -For more tips and tricks to improve your chemistry studies, you can visit the creator's website and get a free guide titled 'Twelve Secrets to Acing Chemistry' by going to www.chemsecrets.com. There is also a link provided in the description of the video.

Outlines
00:00
🧪 Understanding Molality in Chemistry

The first paragraph of the video script introduces the concept of molality, a unit of concentration similar to molarity but independent of temperature. It explains that molality is calculated using the formula moles of solute divided by kilograms of solvent. Two examples are provided to illustrate the calculation process. The first example involves calculating the molality of a KBR solution in water, where 128 grams of KBR is dissolved in 925 grams of water. The process involves converting grams to moles for the solute and grams to kilograms for the solvent, then applying the formula to get a molality of approximately 1.16 mol/kg. The second example is more complex, involving the calculation of molality for a sulfuric acid solution in water at 80 degrees Celsius, taking into account the density of water at that temperature. The process includes converting grams of H2SO4 to moles and milliliters of water to kilograms using the density, resulting in a molality of approximately 0.27 mol/kg. The paragraph concludes with an explanation of how to use density in calculations and emphasizes the importance of converting to kilograms before applying the molality formula.

05:00
📚 Accessing Study Resources for Chemistry Success

The second paragraph shifts focus from the scientific explanation to providing a resource for students looking to improve their chemistry studies. The speaker invites viewers to visit their website to obtain a free guide titled 'Twelve Secrets to Acing Chemistry'. This guide presumably contains strategies, tips, and tricks to excel in chemistry exams and classes. The speaker expresses confidence in the guide's ability to assist students and encourages them to check it out through a link provided in the description below the video. The paragraph ends with words of encouragement for viewers to keep working hard and continue their efforts in their chemistry studies.

Mindmap
Keywords
💡Molality
Molality is a unit of concentration that measures the amount of solute in a solution, expressed in moles per kilogram of solvent. It is a key concept in the video as it is the main topic being discussed. Molality is temperature independent, which means it remains constant regardless of changes in temperature, unlike molarity. In the script, molality is calculated using the formula: moles of solute divided by kilograms of solvent, as demonstrated in the examples provided.
💡Molarity
Molarity is another unit of concentration that measures the amount of solute per volume of solution, typically expressed in moles per liter (mol/L). It is mentioned in the script as a point of comparison to molality, highlighting that molarity is temperature dependent and changes with temperature variations. This contrast is important for understanding when to use molality versus molarity in different chemical contexts.
💡Solute
A solute is the substance that is dissolved in a solvent to form a solution. In the context of the video, the solute is the substance that is being measured in moles for the calculation of molality. The script identifies KBR and H2SO4 as solutes in the examples provided, where their respective masses are converted into moles for the molality calculation.
💡Solvent
A solvent is the substance in which the solute is dissolved to form a solution. In the video, water is consistently used as the solvent. The mass of the solvent, in kilograms, is a necessary component in the molality calculation, as shown in the examples where the mass of water is used to determine the molality of the KBR and H2SO4 solutions.
💡Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is used in the script to convert the mass of the solute from grams to moles, which is a crucial step in calculating molality. The molar mass of KBR and H2SO4 are used in the examples to perform this conversion.
💡Density
Density is a measure of mass per unit volume, often expressed in grams per milliliter (g/mL) or grams per liter (g/L). In the video, the density of water at 80 degrees Celsius is given as 0.997 g/mL. This value is used to convert the volume of water from milliliters to grams, which is then converted to kilograms for the molality calculation in the second example.
💡Conversion
Conversion in the context of the video refers to the process of changing the units of measurement for the solute and solvent to match the requirements of the molality formula. This includes converting grams to moles using molar mass and converting grams to kilograms by dividing by 1000. The script demonstrates these conversions in the examples provided.
💡Significant Figures
Significant figures are the digits in a number that carry meaningful information about its precision. In the script, the concept of significant figures is mentioned in relation to rounding the final answer for molality to two significant figures, which corresponds to the precision of the initial measurement of 15 grams of H2SO4.
💡Chemistry
Chemistry is the scientific study of the composition, structure, properties, and reactions of matter. The video is educational and focused on teaching a specific concept within chemistry—molality. The script aims to help viewers understand and calculate molality, which is an essential skill in the study of chemistry.
💡Study Tips
Study tips are strategies or advice given to help learners improve their understanding and retention of information. In the video, the host offers a free guide called 'Twelve Secrets to Acing Chemistry' and encourages viewers to visit their website for additional study tips and resources to excel in chemistry.
Highlights

Introduction to molality as a unit of concentration.

Difference between molality and molarity, emphasizing molality's temperature independence.

Molality equation: moles of solute divided by kilograms of solvent.

Example 1: Calculating molality of a KBR solution.

Identifying solute and solvent in a solution.

Conversion of solute mass to moles using molar mass.

Conversion of solvent mass to kilograms.

Plugging values into the molality equation.

Example 2: Calculating molality with density involved.

Identifying solute and solvent in a more complex scenario.

Conversion of grams to moles for H2SO4.

Using density to convert milliliters of water to kilograms.

Calculating molality with density and molar mass.

Importance of significant figures in calculations.

Promoting a free guide for chemistry study tips and exam strategies.

Invitation to visit the website for additional chemistry resources.

Encouragement to continue working hard in chemistry studies.

Transcripts
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