What would be the freezing point of 3.00m aqueous solution of a nonvolatile , nonelectrolyte material? The cryoscopic constant of watet is kf=1.86°C/molal
To calculate the freezing point depression of the solution, we can use the formula:
ΔTf = Kf × m × i
where ΔTf is the freezing point depression, Kf is the cryoscopic constant of water, m is the molality of the solute, and i is the van't Hoff factor. Since the solute is a nonvolatile, nonelectrolyte material, the van't Hoff factor i is equal to 1.
Given the values provided:
Kf = 1.86°C/molal
m = 3.00 molal
i = 1
Plugging these values into the formula:
ΔTf = (1.86°C/molal) × (3.00 molal) × 1 = 5.58°C
Now, we know that the freezing point of pure water is 0°C. To find the freezing point of the aqueous solution, we subtract the freezing point depression from the freezing point of pure water:
Freezing point of solution = Freezing point of pure water - ΔTf
Freezing point of solution = 0°C - 5.58°C = -5.58°C
So, the freezing point of the 3.00 molal aqueous solution of a nonvolatile, nonelectrolyte material is -5.58°C.