Which solution has the lower freezing point?
90.0 g CH3OH in 100. g H2O
180.0 g CH3CH2OH in 200. g H2O
I have worked this problem for someone within the last 3-4 days.
delta T = kf m
kf is not needed for this problem.
change 90 g CH3OH and 180 g CH3CH2OH to mols. Change 100 gH2O to kg (0.1 kg) and 200 g H2O to kg (0.2 kg).
Then m = molality = mols/kg.
The one with the largest delta T will have the lower freezing point.
To compare the freezing points of the two solutions, we need to calculate the molality (mols/kg) of each solute.
1. Let's start with 90.0 g of CH3OH (methanol) in 100. g of H2O (water).
- Convert the mass of CH3OH to moles by dividing by its molar mass.
- Molar mass of CH3OH = 32.04 g/mol
- Moles of CH3OH = 90.0 g / 32.04 g/mol = 2.806 mol
- Convert the mass of H2O to kilograms by dividing by 1000.
- Mass of H2O = 100. g / 1000 = 0.1 kg
- Molality of the CH3OH solution = 2.806 mol / 0.1 kg = 28.06 mol/kg
2. Now let's calculate the molality for 180.0 g of CH3CH2OH (ethanol) in 200. g of H2O.
- Convert the mass of CH3CH2OH to moles by dividing by its molar mass.
- Molar mass of CH3CH2OH = 46.07 g/mol
- Moles of CH3CH2OH = 180.0 g / 46.07 g/mol = 3.907 mol
- Convert the mass of H2O to kilograms by dividing by 1000.
- Mass of H2O = 200. g / 1000 = 0.2 kg
- Molality of the CH3CH2OH solution = 3.907 mol / 0.2 kg = 19.54 mol/kg
3. Now, compare the molality values of the two solutions.
- The solution with the higher molality will have the larger change in freezing point (delta T), and consequently, the lower freezing point.
- Comparing the molality values we just calculated, we can see that the CH3OH solution has a higher molality (28.06 mol/kg) compared to the CH3CH2OH solution (19.54 mol/kg).
- Therefore, the CH3OH solution has the lower freezing point of the two solutions.