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# Freezing Point Depression of Water by Sucrose

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We need to calculate the molar mass of sucrose from the freezing point depression of water using a MeasureNet temp probe to get the cooling curves of water and sucrose. We haven't covered "molality" yet, and I'm looking for a simple way to determine these values? I'm more than willing to do the work myself, I just need a push in the right direction, considering we haven't learned this yet. Thanks!

https://brainmass.com/chemistry/stoichiometry/freezing-point-depression-water-sucrose-539150

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Hello!

Molality is a measure of concentration where you have the solute (sucrose) over the solvent (water). Specifically, molality covers moles of solute over kilograms of solvent. If you are familiar with molarity, you will see it's similar, except we're using mass of the solvent instead of volume. This helps us when determining temperature-dependent effects, because volume can change with temperature!

The basic formula you need to use to determine the freezing-point depression is going to be:
dT = Kf * b * i

Kf is the cryoscopic constant, which is constant for a given solvent. Water is 1.853 K*kg/mol
b is the molality of the solute
i is the van't hoff factor, the number of particles you get when you dissolve something. Sucrose stays as one particle when it dissolves (as opposed to NaCl, which breaks up into 2 ions!), so "i" will be equal to 1

Basically, what you're going to be able to do is determine the molality of the sucrose solution using the cooling curve. You'll find the depression of freezing point and get "b" in the equation above.

b = moles solute/kg solvent

You will already know a few things. First, you will have the grams of sucrose you put in, and you will also have the amount of water used. So if you find molality, you will be able to find moles of sucrose. From moles of sucrose, you use the molecular weight formula:

moles of sucrose = grams of sucrose/molecular weight

This will allow you to calculate molecular weight!

Please let me know if I can clarify any of this information for you! Have a great day.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!