Calculating Boiling Point Elevation for a Solution with Dissociating Solutes

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To calculate the boiling point elevation of a solution containing 1.25 mol of CaCl2 in 1400g of water, it's essential to consider that CaCl2 dissociates into three ions: one calcium ion and two chloride ions. The effective molality is determined using the Van't Hoff factor, which for CaCl2 is 3, but the confusion arises from the calculation of the molality itself. The correct molality of the solution is 0.893 molal, not 1.75 m. Using the boiling point elevation equation ΔT = imK_b with the proper values will yield the accurate boiling point elevation. Proper understanding of these concepts is crucial for accurate calculations in colligative properties.
philistinesin
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Here's the question:

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

Now, I know that I have to take into account the fact that CaCl_2 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 \Delta T = imK_b

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.
 
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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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