Week 3.2: Molarity Lab
Nicole Houchins
Created on September 4, 2024
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Molarity & the Chemistry of Kool-Aid Lab
Welcome to Week 3!
- Watch Lecture 3 and complete the discussion post
- Week 3 Content quiz is now open
- If you didn't get 100% on the Syllabus Quiz or Lab Safety Quiz, please retake them until you get 100%!
- Assignments due on Sunday, 9/8 at 11:59pm
Lecture Reminders
- Two post-labs to complete this week!
- Assignments due on Sunday, 9/8 at 11:59pm
Lab Reminders
Objectives
- Define molarity and its associated terms
- Calculate the appropriate amounts and proportions of solvent and solute to make varied molar concentration solutions
- Use laboratory equipment to analyze varied molar concentrations
- Describe and practice safety in a laboratory setting
- Closed-toe shoes
- Goggles
- Gloves
Week 3: Molarity & the Chemistry of Kool-Aid Lab
2. What is the molar mass of the drink mix (C12H22O11)?3. How much (mass) of powdered drink mix is required to make 0.1 L of 0.05 M solution?4. How much (mass) of powdered drink mix is required to make 0.1 L of 0.2 M solution?5. How much (mass) of powdered drink mix is required to make 0.1 L of 0.7 M solution?
Pre-Lab
2. What is the molar mass of the drink mix (C12H22O11)?(12 carbons * 12 g) + (22 hydrogens * 1 g) + (11 oxygens * 16 g) = 342 grams3. How much (mass) of powdered drink mix is required to make 0.1 L of 0.05 M solution?4. How much (mass) of powdered drink mix is required to make 0.1 L of 0.2 M solution?5. How much (mass) of powdered drink mix is required to make 0.1 L of 0.7 M solution?
Pre-Lab
2. What is the molar mass of the drink mix (C12H22O11)?(12 carbons * 12 g) + (22 hydrogens * 1 g) + (11 oxygens * 16 g) = 342 grams3. How much (mass) of powdered drink mix is required to make 0.1 L of 0.05 M solution?0.05 M * 0.1 L * 342 g = 1.71 grams4. How much (mass) of powdered drink mix is required to make 0.1 L of 0.2 M solution?5. How much (mass) of powdered drink mix is required to make 0.1 L of 0.7 M solution?
Pre-Lab
2. What is the molar mass of the drink mix (C12H22O11)?(12 carbons * 12 g) + (22 hydrogens * 1 g) + (11 oxygens * 16 g) = 342 grams3. How much (mass) of powdered drink mix is required to make 0.1 L of 0.05 M solution?0.05 M * 0.1 L * 342 g = 1.71 grams4. How much (mass) of powdered drink mix is required to make 0.1 L of 0.2 M solution?0.2 M * 0.1 L * 342 g = 6.84 grams5. How much (mass) of powdered drink mix is required to make 0.1 L of 0.7 M solution?
Pre-Lab
2. What is the molar mass of the drink mix (C12H22O11)?(12 carbons * 12 g) + (22 hydrogens * 1 g) + (11 oxygens * 16 g) = 342 grams3. How much (mass) of powdered drink mix is required to make 0.1 L of 0.05 M solution?0.05 M * 0.1 L * 342 g = 1.71 grams4. How much (mass) of powdered drink mix is required to make 0.1 L of 0.2 M solution?0.2 M * 0.1 L * 342 g = 6.84 grams5. How much (mass) of powdered drink mix is required to make 0.1 L of 0.7 M solution?0.7 M * 0.1 L * 342 g = 23.94 grams
Pre-Lab