What mass of salt (NaCl) should you add to 1.52 L of water in an ice cream maker to make a solution that freezes at -12.4 ^{o}C ? Assume complete dissociation of the NaCl and density of 1.00 g/mL for water.

We’re being asked to **determine the mass of salt(NaCl)** that must be added to **1.52 L of water** to get a solution that **freezes at –12.4 ˚C**.

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)

Recall that the ** molality of a solution** is given by:

$\overline{){\mathbf{molality}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

**For this problem, we need to do the following:**

* Step 1:* Calculate for ΔT

* Step 2:* Determine the molality of the solution.

* Step 3:* Calculate the mass of NaCl needed.

Freezing Point Depression

Freezing Point Depression

Freezing Point Depression

Freezing Point Depression