You are asked to prepare a temperature bath that freezes at -25 degrees C, using only water and ammonium sulfate.
a. How many grams of ammonium sulfate would you need to prepare this bath if you start with 500 grams of water?
b. What would the boiling point (degrees C) of this solution be?
delta T = i*Kf*m
Plug in delta T, i = 2, and Kf and solv for m.
m = mols/kg solvent
Substitute and solve for mols.
mol = grams/molar mass. You know molar mass and mols, solve for grams.
To answer these questions, we need to understand the concept of freezing point depression and boiling point elevation. Freezing point depression describes the phenomenon where the freezing point of a solvent is lowered when a solute is added, while boiling point elevation refers to the increase in boiling point due to the addition of a solute.
In this case, we are adding ammonium sulfate as the solute to water as the solvent. The freezing point of the resulting solution will be -25 degrees C, which is lower than the freezing point of pure water (0 degrees C). The boiling point of the solution will also be higher than the boiling point of pure water (100 degrees C).
To calculate the amount of ammonium sulfate needed (a), we need to use the formula for molality (moles of solute per kilogram of solvent). The equation is:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point temperature,
Kf is the cryoscopic constant for water (1.86 °C·kg/mol),
m is the molality of the solution in mol solute/kg solvent
First, we convert the given mass of water to kilograms by dividing by 1000:
mass_water = 500 g = 0.5 kg
Since we know the change in freezing point temperature (ΔTf = -25 °C) and the cryoscopic constant (Kf = 1.86 °C·kg/mol), we can rearrange the equation to solve for the molality (m):
m = ΔTf / Kf
m = -25 °C / 1.86 °C·kg/mol
m ≈ -13.44 mol of solute/kg of solvent
Since the molality is calculated based on the solvent (water), we need to determine the amount of ammonium sulfate (NH4)2SO4 needed to provide that molality.
The molar mass of ammonium sulfate is:
molar_mass_ammonium_sulfate = (2 * (1.01 g/mol) + 14.01 g/mol + 4 * (16.00 g/mol)) + 32.07 g/mol
= 132.14 g/mol
To calculate the amount of ammonium sulfate needed, we can use the formula:
mass_ammonium_sulfate = m * mass_water * molar_mass_ammonium_sulfate
mass_ammonium_sulfate = -13.44 mol/kg * 0.5 kg * 132.14 g/mol
mass_ammonium_sulfate ≈ -8,861.2 g
However, since the mass of a substance cannot be negative, we take the absolute value of the calculated mass:
mass_ammonium_sulfate ≈ 8,861.2 g (approximately 8,861 grams)
Therefore, you would need approximately 8,861 grams of ammonium sulfate to prepare the -25 degrees C temperature bath using 500 grams of water.
To find the boiling point (b) of the solution, we can use the equation for boiling point elevation:
ΔTb = Kb * m
Where:
ΔTb is the change in boiling point temperature,
Kb is the ebullioscopic constant for water (0.512 °C·kg/mol),
m is the molality of the solution in mol solute/kg solvent
We can use the previously calculated molality (m = -13.44 mol/kg) and the ebullioscopic constant (Kb = 0.512 °C·kg/mol) to find the change in boiling point temperature:
ΔTb = Kb * m
ΔTb ≈ 0.512 °C·kg/mol * -13.44 mol/kg
≈ -6.88512 °C
To find the boiling point of the solution, we add the change in boiling point temperature to the boiling point of pure water (100 °C):
boiling_point_solution = 100 °C + (-6.88512 °C)
≈ 93.11588 °C
Therefore, the boiling point of this solution would be approximately 93.12 degrees Celsius.