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Lab 7 – Molality Colligative Properties

Lab 7 – Molality Colligative Properties

Name: [First and Last]

Date: DD/MM/YYY

Experiment #: 7

Title:  Molality Colligative Properties

Purpose:

This experiment aims to explore the concept of molality and colligative properties. By calculating the molality of two salts and investigating their ability to lower the temperature of ice, students will learn about the relationship between solute concentration and freezing point depression. Students will also be able to observe the impacts of the salts on the melting point of ice cream during the preparation of ice cream. Our assignment help will hone your writing prowess for papers that will awe your professors.

Procedure: 

Data/Results/Calculations:

1NaCl     → 1Na+   +   1Cl

1 mole of NaCl solids produces 1mole Na+ 1 mole Cl –   ions

Therefore, in all 2moles of ions Na+ + Cl –   yielded

  1. Calcium Chloride: ion production

1 CaCl2 → 1Ca2+ + 2Cl

1 mole of CaCl2   produces 1 mole of Ca2+   2 moles of Cl –   ions

Therefore, in all 3 moles of ions   Ca2+ + Cl –    are produced

  1. Molarity of Sodium Chloride

24g NaCl X1 mol NaCl/58g X 2moles/1moleNaCl = 1.0 mol NaCl ions

0.5mol NaCl  →   0.50mol Na+   + 0.50mol Cl –  

So 1 mol ions in total are produced (0.5 mol + 0.5 mol)

1 mol NaCl ions /0.25 = 4m NaCl

Where “m” is molal

  1. Molarity of Calcium Chloride

55.5g CaCl2 X 1 mole CaCl2  / 111g X  3 mol ions/ 1 mol CaCl2 = 1.5 mol CaCl2 ions

0.5 mol CaCl2  →   0.5 mol Ca2+  + 1.0 mol Cl –  

So 1.5 mol ions in total are produced (0.5 mol + 1.0 mol)

1.5 mole CaClions / 0.25kg = 6 mCaCl2

Where “m” is molal

  1. Molarity of Sodium Chloride V Molarity of Calcium Chloride

The amount of ions generated determines the molarity of a substance

NaCl

1 mole NaCl ions / 0.25 kg = 4m NaCl

CaCl2

1.5 mole CaCl2 ions / 0.5 kg = 6 mCaCl2

Temperature depression NaCl v CaCl2

Temperature depression: change in temperature = kf   X m

Where kf is molal freezing point depression constant (1.86 o C/m for water)

NaCl

Change in temperature = kf   X m

= 1.86 o C/ m X 4m

= 7.44 o C

CaCl2

Change in temperature = kf   X m

= 1.86 o C/ m X 6m

= 11.2 o C

Conclusions:

Through this experiment, the purpose was successfully met as students gained an understanding of molality and colligative properties. They learned how to calculate molality and observed the effects of different salts on the freezing point of ice. Additionally, they discovered the varying impacts of the salts on the melting point of ice cream, highlighting the practical applications of colligative properties in food preparation.

Notes:

Colligative properties, such as freezing point depression and osmotic pressure, are crucial in ice cream production. They help to lower the freezing point of the ice cream mixture, resulting in a smoother texture and preventing the formation of large ice crystals. This enhances the overall quality and palatability of the final product.

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Question 


Lab 7 – Molality Colligative Properties

This lab will be conducted by an experienced instructor. As you watch the lab, you should keep a lab notebook just as you would if you were personally conducting the lab. A well-kept lab notebook is the key to successful labs. The lab notebook acts as the record of your experiment, helps you organize your thoughts and understand the experiment’s results and is useful for writing your lab report. Your lab notebook and/or lab report can be used as you take the lab exam that accompanies each experiment.

To complete the lab report requirement, be sure to follow the “Lab Report Sample” as your model and type out your lab report using the “Lab Report Template”. These can be found under “Lab Overview”. You can upload your report by clicking “Submit,” and then attaching your document.

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