Answer :
Given the osmotic pressure and the freezing-point constant of water, we can calculate the freezing point of the solution to be 68.68 K.
The osmotic pressure of an aqueous solution is related to its concentration through the equation π = i * MRT, where π is the osmotic pressure, i is the van't Hoff factor (the number of particles per formula unit), M is the molarity, R is the ideal gas constant, and T is the temperature in Kelvin. In this case, we know the osmotic pressure is 114 kPa, but the molarity is unknown.
However, we can use the density of water to calculate the molality, which is defined as moles of solute per kilogram of solvent. Since the density of water is 1000 kg/m³, the molality (m) will be the same as the molarity (M).
To find the freezing point of the solution, we can use the formula ΔT = -i * Kf * m, where ΔT is the change in temperature, i is the van't Hoff factor, Kf is the freezing-point constant of water, and m is the molality of the solution. The van't Hoff factor for an aqueous solution is typically close to the number of ions present. Since the solution is not specified, we will assume a van't Hoff factor of 1.
Given that the freezing-point constant of water is 1.86 K-kg-mol⁻¹ and the molality of the solution is equal to its molarity, we can rearrange the formula to solve for ΔT.
Since the freezing point is the temperature at which the solution freezes, ΔT will be negative. Plugging in the values, we have ΔT = -1 * 1.86 * m. Substituting m with 114 kPa (which is equal to 114 mol/kg), we get ΔT = -1 * 1.86 * 114 = -211.32 K.
Finally, to determine the freezing point, we subtract ΔT from the original temperature. The original temperature is 280 K, so the freezing point of the solution will be 280 - 211.32 = 68.68 K.
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