To calculate the freezing point of a solution containing glucose (C₆H₁₂O₆) dissolved in water, we start with the formula for the freezing point depression:

Freezing Point of Solution = Freezing Point of Solvent - ΔT_f

In this case, the solvent is water, which has a freezing point of 0 degrees Celsius. To find ΔT_f, we use the equation:

ΔT_f = i × K_f × m

Here, i is the van 't Hoff factor, which is 1 for glucose since it is a covalent compound and does not dissociate into ions. The freezing point constant (K_f) for water is 1.86 °C/m.

Next, we need to calculate the molality (m) of the glucose solution. First, we convert the mass of water from grams to kilograms:

302.6 g of water = 0.3026 kg of water

Now, we calculate the number of moles of glucose using its molar mass (180.156 g/mol):

Moles of glucose = 110.7 g × (1 mol / 180.156 g) = 0.6145 mol

Now we can find the molality:

m = moles of solute / kg of solvent = 0.6145 mol / 0.3026 kg ≈ 2.03 mol/kg

Now we can substitute the values into the ΔT_f equation:

ΔT_f = 1 × 1.86 °C/m × 2.03 mol/kg ≈ 3.78 °C

Finally, we can find the new freezing point of the solution:

Freezing Point of Solution = 0 °C - 3.78 °C = -3.78 °C

Thus, the freezing point of the glucose solution is approximately -3.78 degrees Celsius.