At –11.0 °C (a common temperature for household freezers), what is the maximum mass of aspartame (C14H18N2O5) you can add to 1.50 kg of pure water and still have the solution freeze? Assume that aspartame is a molecular solid and does not ionize when it dissolves in water. Kf of water is 1.86 degree c/m.
delta T = Kf*m
Substitute and solve for m
m = mols/kg solvent
Substitute and solve for mols.
mols = grams/molar mass.
Substitute and solve for grams.
To determine the maximum mass of aspartame that can be added to 1.50 kg of pure water while still keeping the solution frozen at -11.0 °C, we need to calculate the freezing point depression caused by the addition of aspartame and use it to find the maximum mass.
The freezing point depression (∆Tf) is given by the formula:
∆Tf = Kf * m
where Kf is the freezing point depression constant (1.86 °C/m) and m is the molality of the solution.
First, we need to find the molality (m) of the solution. Molality is defined as the moles of solute per kilogram of solvent.
To determine the number of moles of aspartame, we need to convert the mass of aspartame to moles:
Mass of aspartame (C14H18N2O5) = molecular weight of aspartame * moles of aspartame
Molecular weight of aspartame (C14H18N2O5) = (14 * 12.01 g/mol) + (18 * 1.01 g/mol) + (2 * 14.01 g/mol) + (5 * 16.00 g/mol) = 294.3 g/mol
The molecular weight of aspartame is 294.3 g/mol.
Given that 1 kg of water is equivalent to 1000 g of water, we can calculate the number of moles of aspartame using the following equation:
moles of aspartame = mass of aspartame / molecular weight of aspartame
Next, we need to calculate the molality using the equation:
molality (m) = moles of aspartame / mass of water in kg
Now, we have all the necessary information to calculate the freezing point depression:
∆Tf = Kf * m
Finally, we can solve for the maximum mass of aspartame that can be added while still freezing at -11.0 °C:
mass of aspartame = moles of aspartame * molecular weight of aspartame
Follow the steps mentioned above to calculate the maximum mass of aspartame that can be added.
To solve this problem, we need to use the concept of freezing point depression and the equation:
ΔT = Kf * m * i,
where ΔT is the change in the freezing point, Kf is the cryoscopic constant, m is the molality of the solute, and i is the van't Hoff factor.
First, we need to calculate the change in freezing point by using the equation:
ΔT = -11.0 °C - 0 °C = -11.0 °C.
Next, we need to calculate the molality of the solute. Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is aspartame (C14H18N2O5), and the solvent is water.
First, let's calculate the molar mass of aspartame:
C = 12.01 g/mol
H = 1.01 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of aspartame (C14H18N2O5) = (12.01 * 14) + (1.01 * 18) + (14.01 * 2) + (16.00 * 5) = 294.31 g/mol.
Now, we can calculate the number of moles of aspartame needed to achieve the freezing point depression. Let's assume x moles of aspartame.
Mass of aspartame (m) = x moles * 294.31 g/mol.
To calculate the molality (m), we divide the moles of aspartame by the mass of the water in kg:
m = (x moles * 294.31 g/mol) / 1.50 kg.
Now, let's substitute the values into the freezing point depression equation:
ΔT = Kf * m * i.
-11.0 °C = (1.86 °C/m) * [(x moles * 294.31 g/mol) / 1.50 kg] * i.
Since aspartame does not ionize in water, the van't Hoff factor (i) is 1.
-11.0 °C = (1.86 °C/m) * [(x moles * 294.31 g/mol) / 1.50 kg].
Now, we can solve for x moles of aspartame:
x moles = (-11.0 °C * 1.50 kg * 1 mol/294.31 g) / (1.86 °C/m).
x moles = -7.0047 mol.
Finally, to calculate the maximum mass of aspartame, we multiply the number of moles by the molar mass:
Mass of aspartame = -7.0047 mol * 294.31 g/mol.
Therefore, the maximum mass of aspartame you can add to 1.50 kg of water and still have the solution freeze at -11.0 °C is approximately 2066.53 g.