Freezing-Point Depression — Glucose in Water
By Prof. Marco Ruzzi.
Theoretical background
Pure deionized water freezes at 0 °C under the conditions considered here. A dissolved solute lowers its freezing point. For the glucose solution:
\[|\Delta T|=K_{\mathrm{cr}}b.\]The cryoscopic constant Kcr has units °C kg mol⁻¹. Molality is the amount of solute divided by the mass of solvent in kilograms:
\[b=\frac{n_{\mathrm{solute}}}{m_{\mathrm{solvent}}}.\]Exercise 8 — Freezing point of a glucose solution
What is the freezing point of a solution containing 25.0 g of glucose, C₆H₁₂O₆, in 250.0 g of water?
Use Kcr=1.86 °C kg mol⁻¹ and a glucose molar mass of 180.157 g mol⁻¹.
1. Amount of glucose
\[n_{\mathrm{glucose}}=\frac{25.0}{180.157}\ \mathrm{mol}.\] \[n_{\mathrm{glucose}}\simeq0.13877\ \mathrm{mol}.\]2. Molality
The solvent mass is 250.0×10⁻³ kg:
\[b=\frac{0.13877}{0.2500}\ \mathrm{mol\,kg^{-1}}.\] \[b\simeq0.5551\ \mathrm{mol\,kg^{-1}}.\]3. Freezing-point depression
\[|\Delta T|=(1.86)(0.5551)\ ^\circ\mathrm{C}.\] \[|\Delta T|\simeq1.0325\ ^\circ\mathrm{C}.\]The freezing point is lowered by this amount, rather than raised:
\[T_f\simeq-1.03\ ^\circ\mathrm{C}.\]