What is the freezing point of a 10% (by mass) solution of CH3OH in water?
10% w/w means 10g solute/100 g solution. That is 10g CH3OH/(90g H2O + 10g CH3OH).
Convert 10 g CH3OH to mols. That's mols = grams/molar mass = ?
Change to m. m = mols/kg solvent. You have mols and kg solvent is 0.025 g.
Then delta T = Kf*molality
You have Kf (1.86), m from above, solve for delta T.
Subtract delta T from the normal freezing point (0C) to find the the new freezing point.
To determine the freezing point of a solution, we need to use the equation for freezing point depression, which is a colligative property. The equation is given by:
ΔT_f = K_f * molality
Where:
ΔT_f is the freezing point depression (in °C)
K_f is the cryoscopic constant (specific to the solvent; for water, it is 1.86 °C/m)
molality is the molal concentration of the solute (in mol solute/kg solvent)
In this case, we have a 10% (by mass) solution of CH3OH in water. To calculate the molality, we need to convert the mass percent into molal concentration.
Step 1: Calculate the mass of CH3OH
Assume we have 100 g of the solution, which means we have 10 g of CH3OH (since it is a 10% solution by mass).
Step 2: Convert mass of CH3OH to moles
To convert grams to moles, we need to divide the mass by the molar mass of CH3OH, which is approximately 32 g/mol.
moles of CH3OH = 10 g / 32 g/mol ≈ 0.3125 mol
Step 3: Calculate the molality
Molality is defined as the moles of solute divided by the mass of the solvent (in kg).
mass of water = 100 g (since we assumed 100 g of the solution)
mass of water = 0.1 kg
molality = 0.3125 mol / 0.1 kg ≈ 3.125 mol/kg
Step 4: Calculate the freezing point depression
Using the freezing point depression equation, with K_f = 1.86 °C/m and molality = 3.125 mol/kg:
ΔT_f = 1.86 °C/m * 3.125 mol/kg
ΔT_f ≈ 5.81 °C
Step 5: Calculate the freezing point of the solution
The freezing point depression is the difference between the freezing point of the pure solvent (water) and the freezing point of the solution.
Freezing point of pure water = 0 °C
Freezing point of solution = 0 °C - 5.81 °C = -5.81 °C
Therefore, the freezing point of a 10% (by mass) solution of CH3OH in water is approximately -5.81 °C.
To find the freezing point of a solution, you need to use the concept of freezing point depression, which states that the freezing point of a solution is lower than that of the pure solvent. This is due to the presence of a solute particles disrupting the formation of a crystal lattice.
To calculate the freezing point depression, you need to know the molality (moles of solute per kilogram of solvent) and the cryoscopic constant (Kf) of the solvent.
In this case, we have a 10% (by mass) solution of CH3OH (methanol) in water. To convert the mass percentage to molality, we need to know the density of the solution or assume an arbitrary mass of the solution. For simplicity, let's assume we have 100 grams of the solution.
Since we're working with 100 grams of the solution, we have 10 grams of CH3OH and 90 grams of water. To calculate the molality, we need to convert the grams of CH3OH and water to moles.
1. Calculate the moles of CH3OH:
- Find the molar mass of CH3OH: (1 * 12.01) + (4 * 1.01) + 16.00 = 32.05 g/mol
- Convert grams to moles: 10 g * (1 mol / 32.05 g) = 0.312 mol
2. Calculate the moles of water:
- Find the molar mass of H2O: 2 * 1.01 + 16.00 = 18.02 g/mol
- Convert grams to moles: 90 g * (1 mol / 18.02 g) = 4.99 mol
3. Calculate the molality (moles of solute per kilogram of solvent):
- molality (m) = moles of solute / kilograms of solvent
- molality (m) = 0.312 mol / 0.090 kg = 3.47 m
Now, we need to know the cryoscopic constant (Kf) for water. The cryoscopic constant for water is approximately 1.86 °C•kg/mol.
Finally, we can calculate the freezing point depression using the formula:
ΔTf = Kf * m
where:
ΔTf = Freezing point depression
Kf = Cryoscopic constant
m = Molality
ΔTf = 1.86 °C•kg/mol * 3.47 m = 6.45 °C
Therefore, the freezing point of the 10% solution of CH3OH in water is 0°C - 6.45°C = -6.45°C.