I have the answer but how do you solve for this problem:
What is the freezing point of a sugar solution prepared by dissolving 375.00g of sucrose (342.3g/mol) in 1500.0g of water (Kf of water=1.86 degrees Celcius)?
To solve for the freezing point of the sugar solution, you need to use the formula for calculating the freezing point depression. The freezing point depression (∆Tf) is calculated by multiplying the molality (m) of the solution by the freezing point constant (Kf) of the solvent.
1. Calculate the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
First, convert the mass of the solute (sucrose) from grams to moles:
Moles of sucrose = (mass of sucrose) / (molar mass of sucrose)
Moles of sucrose = 375.00g / 342.3g/mol
Next, convert the mass of the solvent (water) from grams to kilograms:
Mass of water = 1500.0g / 1000g/kg
Now, calculate the molality:
Molality (m) = (moles of sucrose) / (mass of water in kg)
2. Multiply the molality (m) by the freezing point constant (Kf) of water:
∆Tf = (m) x (Kf)
Given that the Kf of water is 1.86 degrees Celsius (Kf = 1.86°C/kg*mol) in this case.
3. Finally, subtract the calculated freezing point depression (∆Tf) from the freezing point of pure water to find the freezing point of the sugar solution:
Freezing point of solution = Freezing point of pure water - ∆Tf
The freezing point of pure water is 0°C.
By following these steps, you should be able to solve for the freezing point of the sugar solution.