We’re being asked to calculate the mass of Ca3(PO4)2 needed to achieve the freezing point of the solution as the one given with sucrose.
Recall that the freezing point of a solution is lower than that of the pure solvent and the change in freezing point (ΔTf) is given by:
The change in freezing point is also related to the molality of the solution:
where:
i = van’t Hoff factor
m = molality of the solution (in m or mol/kg)
Kf = freezing point depression constant (in ˚C/m) of the solvent
We go through the following steps to solve the problem:
Step 1. Calculate the molality of the solution
Step 2. Calculate the moles of the solute
Step 3. Calculate the mass of the solute
A solution is prepared by dissolving 456 g of sucrose (C12H22O11, molar mass 342 g/mol) in 764 g of water (18.02 g/mol; density 1.00 g/mL) calculate:
d) How many grams of Ca 3 (PO4)2 must be added to 764g of water to have the same effect?
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Our tutors have indicated that to solve this problem you will need to apply the The Colligative Properties concept. You can view video lessons to learn The Colligative Properties. Or if you need more The Colligative Properties practice, you can also practice The Colligative Properties practice problems.