At –14.5 °C (a common temperature for household freezers), what is the maximum mass of aspartame (C14H18N2O5) you can add to 3.00 kg of pure water and still have the solution freeze
To find the maximum mass of aspartame that can be added to 3.00 kg of pure water and still have the solution freeze at -14.5 °C, we need to consider the freezing point depression caused by the presence of the solute.
The formula to calculate the freezing point depression is:
ΔTf = Kf * molality
where ΔTf is the change in freezing point, Kf is the molal freezing point depression constant for water (1.86 °C kg/mol), and molality is the concentration of the solute in mol/kg.
First, we need to convert the mass of water to moles. The molar mass of water (H2O) is 18.015 g/mol.
The number of moles of water = mass of water / molar mass of water
Number of moles of water = 3000 g / 18.015 g/mol ≈ 166.54 mol
Next, we need to calculate the molality of the solution, which is the number of moles of solute (aspartame) per kilogram of solvent (water).
Let's assume the maximum mass of aspartame is m grams.
The molar mass of aspartame (C14H18N2O5) is 294.31 g/mol.
The number of moles of aspartame = mass of aspartame / molar mass of aspartame
Number of moles of aspartame = m g / 294.31 g/mol
Since the molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water to kilograms:
Mass of water = 3.00 kg = 3000 g
Now we can calculate the molality:
molality = (Number of moles of aspartame) / (Mass of water in kg)
molality = (m g / 294.31 g/mol) / (3.00 kg)
To calculate the change in freezing point, we can use the equation:
ΔTf = Kf * molality
ΔTf = (1.86 °C kg/mol) * (molality)
Since the solution should freeze at -14.5 °C, the change in freezing point is:
ΔTf = -14.5 °C - (-0 °C) = -14.5 °C
Finally, solving the equation for the maximum mass of aspartame (m) gives:
m = ΔTf * (molar mass of aspartame) / (Kf * molality)
Substituting the known values:
m = -14.5 °C * 294.31 g/mol / (1.86 °C kg/mol * (m g / 294.31 g/mol) / (3.00 kg)
Simplifying and solving for m will give you the maximum mass of aspartame you can add to the water and still have the solution freeze at -14.5 °C.
delta T = 14.5
delta T = Kf*m
Solve for m
m = mols aspartame/kg solvent
Solve for molsl aspartame
mola = g aspartame/molar mass aspartame
solve for grams.