Question

. An aqueous solution having a density of 1.049 g/cm3 is found to have an osmotic pressure of 17.0 atm at 25°C. Find the temperature at which this solution freezes considering Kf = 1.86 °C/m.

Answer 1

Explanation:

From the osmotic pressure of the solution, calculate the molarity of the solution:

Osmotic pressure = MRT

17 atm = M (0.08206 Latm/molK)(298 K)

M = 0.695 M

Now, to eventually calculate a freezing point, we have to get from molarity to molality of the solution.

First, assume you have 1 L of the solution. That liter contains 0.695 moles of fructose, which has a mass of 0.695 mol X 180.16 g/mol = 125.2 grams

The mass of that 1 L of solution is 1000 cm^3 X 1.049 g/cm^3 = 1049 grams. The mass of water in that solution is 1049 g - 125 g = 924 g H2O

So, the molality of the solution is 0.695 mol / 0.924 kg water = 0.752 molal

The change in the freezing point of that solution is:

Delta Tf = -kf m = -1.86 C/m (0.752 m) = -1.40 degrees C

So, the solution will have a freezing point of -1.40 degrees C.

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Answer 2

Answer:

The aqueous solution will freeze at -32.2°C.

Explanation:

The freezing point of the aqueous solution can be calculated using the freezing point depression equation, which states that the freezing point of a solution is lower than that of the pure solvent and is proportional to the molal concentration of the solute. In this case, the freezing point depression can be calculated by rearranging the equation and substituting the given values as follows:

Freezing Point Depression = (Osmotic Pressure x Kf) / Molality

Freezing Point Depression = (17.0 atm x 1.86°C/m) / 1.049 g/cm3

Freezing Point Depression = 32.2°C

Since the freezing point of pure water is 0°C, the freezing point of the aqueous solution can be calculated by subtracting the freezing point depression from the freezing point of the pure solvent.

Freezing Point = 0°C - 32.2°C

Freezing Point = -32.2°C

Therefore, the aqueous solution will freeze at -32.2°C.

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