how many grams of methanol must be added to 5.00kg of water to lower its freezing point to -12.0 degrees C? for each mole of solute the freezing point of 1kg of water is lowered 1.86 degrees C?
You want delta T to be 12.5 C.
delta T = Kf*m
You know delta T and Kf, solve for molality (m).
m = mols CH3OH/kg solvent. You know kg solvent and m, solve for mols CH3OH.
Then mols CH3OH = grams/molar mass. YOu know molar mass and mols, solve for grams.
Hello DrBob222, thank you for trying to help explain it, but unfortunately it was not helpful.
To calculate the grams of methanol needed to lower the freezing point of water, we can use the formula:
ΔT = K * m
Where:
ΔT is the change in freezing point temperature
K is the cryoscopic constant (molal freezing-point depression constant) for water (1.86 °C·kg/mol)
m is the molality of the solution (moles of solute per kilogram of solvent)
First, we need to calculate the molality of the solution. The molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solvent is water.
The formula to calculate molality is:
molality (m) = moles of solute / mass of solvent (in kg)
Given that the mass of water is 5.00 kg, we can determine the number of moles of solute required to lower the freezing point to -12.0 °C. Since the relationship between change in freezing point and molality is 1:1, the change in freezing point is equal to the molality.
ΔT = m = -12.0 °C
Now we can calculate the moles of solute:
m = moles of solute / 5.00 kg
moles of solute = m * 5.00 kg
Next, we convert the moles of solute to grams of methanol using the molar mass of methanol. The molar mass of methanol (CH3OH) is approximately 32.04 g/mol.
grams of methanol = moles of solute * molar mass of methanol
By substituting the values into the equation, we can find the grams of methanol needed to lower the freezing point of water to -12.0 °C.