At –10.00 °C (a common temperature for household freezers), what is the maximum mass of fructose (C6H12O6) you can add to 1.00 kg of pure water and still have the solution freeze? Assume that fructose is a molecular solid and does not ionize when it dissolves in water.
delta T = Kf*m
Solve for m fructose.
m = moles fructose/kg solvent
Solve for mole fructose
moles = grams/molar mass
You know moles fructose and molar mass fructose, solve for grams fructose.
To determine the maximum mass of fructose you can add to 1.00 kg of pure water in order for the solution to still freeze at -10.00 °C, we need to calculate the freezing point depression caused by the fructose.
The freezing point depression equation is given by:
ΔTf = Kf * molality
Where:
ΔTf is the freezing point depression
Kf is the cryoscopic constant (for water, it is 1.86 °C/m)
molality is the moles of solute per kilogram of solvent
To find the maximum mass of fructose, we need to determine the molality of the solution. First, we convert the mass of water into moles:
1.00 kg of water * (1 mol H2O / 18.01528 g) = 55.495 mol
Since fructose dissolves as a molecular solid and does not ionize, the number of moles of fructose is equal to the number of molecules. Therefore, the molality of the solution is:
molality = moles of fructose / mass of water(in kg)
Next, we rearrange the freezing point depression equation to solve for moles of fructose:
moles of fructose = ΔTf * molality / Kf
Given that the freezing point depression (ΔTf) is -10.00 °C, and the cryoscopic constant (Kf) for water is 1.86 °C/m, we can substitute these values into the equation:
moles of fructose = (-10.00 °C) * (molality) / (1.86 °C/m)
Finally, we convert moles of fructose to grams by multiplying by the molar mass of fructose (C6H12O6), which is approximately 180.16 g/mol:
mass of fructose = moles of fructose * molar mass of fructose
Now, we can calculate the maximum mass of fructose by substituting the values:
mass of fructose = (-10.00 °C) * (molality) / (1.86 °C/m) * (180.16 g/mol)
Note: To calculate an accurate result, the molality should be expressed in moles of fructose per kilogram of water.
Please provide the molality (moles of fructose per kilogram of water) of the solution.
To solve this problem, we need to understand the concept of freezing point depression. When a solute, such as fructose, is added to a solvent, such as water, it lowers the freezing point of the solution compared to the pure solvent. The amount by which the freezing point is lowered depends on the concentration of the solute.
This can be calculated using the equation ΔTf = Kf * m, where ΔTf is the freezing point depression, Kf is the cryoscopic constant (specific for each solvent), and m is the molality of the solution (moles of solute per kilogram of solvent).
In this case, we want the solution to freeze, meaning the freezing point depression should be equal to or greater than zero. The freezing point of pure water is 0.00 °C, and we want to find the maximum mass of fructose that can be added while still allowing the solution to freeze at -10.00 °C.
Now we can calculate the maximum molality of the fructose in the solution using the equation:
ΔTf = Kf * m
Rearranging the equation to isolate m:
m = ΔTf / Kf
Substituting the values:
m = (-10.00 °C - 0.00 °C) / Kf
Since the problem doesn't provide the cryoscopic constant (Kf) for water, we will assume it to be 1.86 °C*kg/mol.
m = (-10.00 °C - 0.00 °C) / 1.86 °C*kg/mol
Now we have the maximum molality that fructose can have in the solution while still allowing it to freeze.
Next, we need to convert the molality to mass. The molar mass of fructose (C6H12O6) is 180.16 g/mol, so we can calculate the maximum mass of fructose using the equation:
Mass of fructose = m * molar mass * mass of water
Substituting the values:
Mass of fructose = m * 180.16 g/mol * 1.00 kg
Now we can calculate the maximum mass of fructose that can be added to 1.00 kg of water while still allowing the solution to freeze at -10.00 °C.
Note that this is a theoretical calculation, and in practice, there may be other factors that could affect the freezing point depression.