Final answer:
The freezing point of water decreases when a solute like lithium hydroxide is dissolved in it, but due to the small quantity of lithium and the low molality of the solution, the freezing point depression is negligible.
Step-by-step explanation:
The freezing point of a solution is affected by the presence of a solute due to a colligative property known as freezing point depression. In particular, lithium reacts with water to form lithium hydroxide (LiOH) and hydrogen gas (H₂). Assuming complete dissociation, each lithium atom produces one Li⁺ ion and one OH⁻ ion. The cube of lithium has a volume of 1.0 mm x 1.0 mm x 1.0 mm, which equals 1.0 x 10⁻¹ cm³. With a density of 0.535 g/cm³, the mass of the lithium cube is 0.535 g/cm³ x 1.0 x 10⁻¹ cm³ = 5.35 x 10⁻⁴ g. The molar mass of lithium is approximately 6.94 g/mol, so the number of moles of Li is (5.35 x 10⁻⁴ g) / (6.94 g/mol) = 7.71 x 10⁻µ mol. The freezing point depression is given by ΔTf = i x Kf x m, where i is the van't Hoff factor (for LiOH, i = 2 since Li⁺ and OH⁻ are produced), Kf is the freezing-point-depression constant for water (1.86°C/m), and m is the molality (moles of solute per kilogram of solvent). The mass of water is 550 g or 0.550 kg. Thus, the molality of the solution is (7.71 x 10⁻µ mol) / (0.550 kg) = 1.40 x 10⁻´ m. Hence, the freezing point depression is ΔTf = 2 x 1.86°C/m x 1.40 x 10⁻´ m = 5.19 x 10⁻´ °C. Since the freezing point of pure water is 0°C, the freezing point of the solution will be slightly lower, but the small amount of salt and the low concentration make this depression negligible for practical purposes.