When 13.62 g (about one tablespoon) of table sugar (sucrose,
C_{12}H_{22}O_{11}) is dissolved in 241.5 mL of water (density 0.997 g/mL), the final volume is 250.0 mL (about one cup).

Would the syrup freeze in a typical freezer (-18 C)? Why or why not?

We’re being asked to **determine whether a syrup freezes at a typical freezer (-18°C)**.

The syrup is prepared by dissolving **13.62 g of table sugar** (sucrose, C_{12}H_{22}O_{11}) in **241.5 mL of water** for which the final volume of the solution is 250.0 mL.

To answer this, we need to **determine the freezing point of the solution** and compare it to the temperature of a typical freezer. When calculating the freezing point of a solution, we’re going to use the equation for Freezing Point Depression.

$\overline{){\mathbf{\u2206}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{i}}{\mathbf{\xb7}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{\xb7}}{\mathbf{m}}}$

∆T_{f} = change in freezing point = T_{f pure solvent} –T_{f solution}

K_{f} = freezing point depression constant

i = van' t Hoff factor of the solute = no. of ions

m = molality

Freezing Point Depression

Freezing Point Depression

Freezing Point Depression

Freezing Point Depression