Experiment VII

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Portage Learning *

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103

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Chemistry

Date

Feb 20, 2024

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docx

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2

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Chemistry Lab Report Name: Date: 06/06/2022 Experiment #: 7 Title: Molality Colligative Properties Purpose: To demonstrate the Molality Colligative Properties of Freezing Point Depression Procedure: We will be using salts NaCl and CaCl 2 to make ice cream 1. Get two large zip lock bags containing ice (1 gallon) and label each bag the name of the salt we will be using 2. Get two large zip lock bags containing mixture of milk, vanilla, and sugar and label each bag the name of the salt we will be using 3. Obtain 0.5 mol of NaCl and CaCl 2 and place them in two separate cups 4. Add 0.5 mol NaCl into the ice bag labeled NaCl and mix it Do the same with 0.5 mol CaCl 2 5. Then put the bag containing the ice cream mixture into each ice bag. 6. Put each bag in a shaker and leave it in there for 15 minutes 7. Data/Results/Calculations: NaCl: ion production 1 NaCl 1 Na + + 1 Cl - 2 moles of ions were produced. Calculation of Molality Molality(m) = mol solute/ kg of solvent 24g NaCl (1 mol NaCl/ 58g) (2 mol ions/ 1 mol NaCl= 1.0 mole NaCl ions 0.5 mol NaCl 0.5 mol Na + + 0.5 mol Cl - m= 1 mol NaCl/ 0.25 kg= 4 m NaCl CaCl 2 : ion production 1 CaCl 2 1 Ca 2+ + 2 Cl - 3 moles of ions were produced
Chemistry Lab Report Calculation of Molality 55.5g of CaCl 2 (1 mole CaCl 2 /111g) (3 mol ions) (1 mol CaCl 2 ) = 1.5 mol CaCl 2 ions 0.5 mol CaCl 2 0.5 mol Ca 2+ + 1 mol Cl - m= 1.5 mol CaCl 2 ions/0.25 kg = 6 m CaCl 2 Temperature Depression Change in T= k f x m k f = 1.86 C/m for H 2 O NaCl Change in T= k f x m Change in T= (1.86 C/m) x 4 m = 7.44 C CaCl 2 Change in T= k f x m Change in T= (1.86 C/m) x 6 m = 11.2 C Conclusions: In this lab we were able to calculate the molality of a salt when dissolve in ice. This technique is used to created ice and, in this lab, we observe and compare salts NaCl and CaCl 2 . Both salts are colligative properties which mean when they were added to and dissolve in ice, they will lower the freezing point temperature. NaCl has a molality of 4 m and change in temperature of 7.44 C. While CaCl 2 has a molality of 6 m and change in temperature of 11.2 C. In the lab we were able to see that CaCl 2 ice cream melted faster than NaCl which proves that NaCl has a higher freezing point. Notes: Salt is a Colligative Property o Colligative property: whenever a dissolve solute into a solvent, it lowers the FP. Molality(m) = mol solute/ kg of solvent Molarity(M)= mol/L
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