Final answer:
The 0.10 m aqueous solutions of glucose and sucrose have the same freezing point depression due to equal molality, thereby predicting identical freezing points for these solutions. This is because both are non-electrolyte solutes that do not ionize and contribute equally to the lowering of the freezing point.
Step-by-step explanation:
The freezing point of a 0.10 m aqueous solution of glucose (180 g/mol) compared to the freezing point of a 0.10 m aqueous solution of sucrose (342 g/mol) can be predicted based on the colligative properties of solutions. The colligative property relevant here is freezing point depression, which holds that the freezing point of a solution is lower than that of the pure solvent. This is due to the presence of solute particles that disrupt the formation of a solid (crystal lattice) from the liquid.
Since glucose and sucrose are both molecular non-electrolytes, they do not dissociate into ions when dissolved in water. The freezing point depression therefore is only dependent on the molality (moles of solute per kilogram of solvent) of the solution and not the type of solute, provided the solute does not ionize. Given that both solutions have the same molality (0.10 m), the freezing point depression for both solutions should be the same. This is assuming that we're comparing solutions with equal molal concentration and that the solutes behave as ideal nonvolatile solutes.
Remember, the greater the concentration of particles, the lower the freezing point will be. However, in this case, since both solutions are 0.10 m, the number of particles per kilogram of water is the same, so the freezing point of both solutions should be lowered equally, making their freezing points the same. Therefore, a 0.10 m aqueous solution of glucose will have an equivalent freezing point as a 0.10 m aqueous solution of sucrose, and both will have a lower freezing point than pure water.