Q.A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucose solution. (269.06 K)

Step 1. In 100 g of 5% cane sugar solution: 5 g sucrose (M=342 g/mol), 95 g water. Moles sucrose $n_2=5/342=0.014620$ mol; molality $m=0.014620/0.095=0.15389$ mol/kg.

Step 2. Taking pure water's freezing point as 273.15 K (0 degC), the sucrose solution's depression is $\Delta T_f = 273.15-271 = 2.15$ K. This gives (via $\Delta T_f=K_fm$) $K_f = 2.15/0.15389 = 13.970$ K kg mol-1 for this particular problem's implied conditions (this is NOT the standard 1.86 value -- it is simply back-calculated from the given data, and is exactly the quantity we now re-use for glucose, since Kf is a solvent property shared by both sugar solutions).

Step 3. In 100 g of 5% glucose solution: 5 g glucose (M=180 g/mol), 95 g water. Moles glucose $n_2'=5/180=0.027778$ mol; molality $m'=0.027778/0.095=0.29247$ mol/kg. …

Use the cane-sugar freezing-point data to back-calculate an implied Kf, then apply that same Kf to th …

• Using 273 K instead of 273.15 K for pure water's freezing point -- this problem's numbers only reproduce the source's stated answer (269.06 K) when the more precise 273.15 K baseline is used consistently in both steps. …