Recall that the melting point is the same temperature as the freezing point.

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:

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

In ice-cream making, the ingredients are kept below 0.0°C in an ice-salt bath.*(a) Assuming that NaCl dissolves completely and forms an ideal solution, what mass of it is needed to lower the melting point of 5.5 kg of ice to –5.0°C?***(b) Given the same assumptions as in part (a), what mass of CaCl _{2} is needed?**

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.