We’re being asked to **calculate the mass of Ca _{3}(PO_{4})_{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 (ΔT_{f})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}{\mathbf{-}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{solution}}}$

The ** change in freezing point** is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{imK}}}_{{\mathbf{f}}}}$

where:

**i** = van’t Hoff factor

**m** = molality of the solution (in m or mol/kg)

**K _{f}** = 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 (C_{12}H_{22}O_{11}, 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 }(PO_{4})_{2 }must be added to 764g of water to have the same effect?

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