Calculating Boiling Point Elevation for a Solution with Dissociating Solutes

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SUMMARY

The boiling point of a solution containing 1.25 mol of CaCl2 in 1400g of water can be calculated using the boiling point elevation equation ΔT = imKb. The Van't Hoff Factor (i) for CaCl2 is 2, as it dissociates into two ions (Ca2+ and 2Cl-). The effective molality of the solution is 0.893 molal, which is derived from the correct application of the molality definition and the dissociation of the solute. This results in an accurate calculation of the boiling point elevation.

PREREQUISITES
  • Understanding of colligative properties, specifically boiling point elevation.
  • Familiarity with the Van't Hoff Factor and its application in solutions.
  • Knowledge of molality and how to calculate it.
  • Basic understanding of ionic dissociation in solutions.
NEXT STEPS
  • Study the calculation of boiling point elevation using different solutes.
  • Learn about the Van't Hoff Factor for various ionic compounds.
  • Explore the concept of molality in detail, including its applications in colligative properties.
  • Investigate the effects of temperature on solubility and boiling point elevation.
USEFUL FOR

Chemistry students, educators, and professionals involved in solution chemistry, particularly those focusing on colligative properties and boiling point calculations.

philistinesin
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Here's the question:

What is the boiling point of a solution that contains 1.25 mol [tex]CaCl_2[/tex] in 1400g of water?

Now, I know that I have to take into account the fact that [tex]CaCl_2[/tex] disassociates in water to form ions.

I thought the effective molality here would be 3 * 1.75 m since there are 3 ions. But, the book saids it's 2 * 1.75 m.

I'm confused as to why that is.
 
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Use your boiling point elevation equation [tex]\Delta T = imK_b[/tex]

The reason why the molality is what it should be has to do with the 'Van't Hoff Factor', i, which is multiplied by the molality of the solution to get the 'effective' molality.
 
Last edited:
Molality, if I'm remembering it correctly, is defined as the mole amount of a solution whose solvent weighs 1000 grams. So the CaCl2 solution you are dealing with is of 0.893 molal. So a recalculation and using this value may give the answer you're looking for.
 

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