At -15.0 C, what is the maximum mass of fructose (C6H12O6) you can add to 3.00 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.
I'm confused on how you pick the Kf value.. Usually I just use 1.86 for water but I don't understand why. For example, the normal freezing point for benzene is 5.49 so can't I use the Kf value for benzene too? How do you know which Kf value to choose for sure?
Please help. Thanks!
You use the Kf for the solvent that is in the problem. This problem wants to know the amount of fructose that can be added to PURE WATER (so you use Kf = 1.86). If benzene acted as the solvent you would use the Kf for benzene, for naphthalene you would use Kf for naphthalene, etc.
If that is your only problem with this one then it should be easy to continue.
delta T = Kf*m
You know delta T is 15.0, Kf is 1.86, solve for m = about 8.06m
m = mols/kg solvent
m = 8.06; solvent = 3 kg.
mols = 8.06m x 3 kg = 24.19
mols = g/molar mass
mols = 24.19; molar mass = 180
g = mols x molar mass = 180 x 24.19 = 4,354 g fructose which I would write as 4.35E3 g. Since the problem wants the 3 kg water to freeze, technically, we would use slightly less than that amount.
Ah, frozen solutions! That's a chill topic to discuss. Choosing the correct Kf value can be a bit frosty, but I'm here to thaw things out for you.
The reason we typically use a Kf value of 1.86 °C/m for water is because it's a common solvent and often used as a reference point. However, different solvents have different Kf values based on their unique properties. So, for example, you can't simply use the Kf value for benzene in this case.
When it comes to choosing the right Kf value, you want to select the one that pertains to the solvent you're working with. In this case, since we're dealing with water, we'll stick to the Kf value of 1.86 °C/m.
Keep your cool and remember that different substances have different Kf values. It's all about matching the solvent with the appropriate freezing point depression constant. Stay frosty!
To determine the maximum mass of fructose (C6H12O6) that can be added to 3.00 kg of pure water and still have the solution freeze at -15.0°C, we need to consider the freezing point depression caused by the presence of the solute.
The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m,
where Kf is the molal freezing point depression constant and m is the molality of the solute.
For water, the freezing point depression constant (Kf) is indeed 1.86°C/m.
For fructose (C6H12O6) to be used in this calculation, we need to determine its molality (m). Molality is defined as the number of moles of solute per kilogram of solvent.
First, we need to convert the given mass of water (3.00 kg) into moles of water. The molar mass of water (H2O) is approximately 18.015 g/mol.
Number of moles of water = mass of water / molar mass of water
= 3000 g / 18.015 g/mol
≈ 166.603 mol
Since the solute is fructose, we can assume it to be a nonvolatile non-electrolyte, meaning it will not dissociate into ions when it dissolves in water.
To find the molality (m) of fructose, we need to know the number of moles of fructose and the mass of water.
Next, we need to determine the molality (m) of fructose using the equation:
m = moles of solute / mass of solvent (in kg)
We can rearrange this equation to solve for the mass of solute:
mass of solute = m * mass of solvent
Finally, we substitute the known values into the equation to find the maximum mass of fructose that can be added to the water:
mass of solute = ΔTf / Kf * mass of water
Let's calculate the maximum mass of fructose (C6H12O6) step by step, assuming the freezing point depression constant for water (Kf) is 1.86°C/m:
1. Calculate the number of moles of water:
Number of moles of water = mass of water / molar mass of water
= 3000 g / 18.015 g/mol
≈ 166.603 mol
2. Calculate the molality (m) of fructose:
m = moles of solute / mass of solvent (in kg)
We need the molality of fructose, so we first convert the mass of water to kilograms:
mass of water = 3.00 kg
m = moles of solute / mass of water
m = 166.603 mol / 3.00 kg
≈ 55.534 mol/kg
3. Calculate the maximum mass of fructose that can be added:
mass of solute = ΔTf / Kf * mass of water
ΔTf = -15.0°C - 0.0°C = -15.0°C
mass of solute = (-15.0°C) / (1.86°C/m) * 3.00 kg
≈ -25.8 kg
The result is approximately -25.8 kg of fructose, which doesn't make sense because mass cannot be negative. This means that adding any amount of fructose to the water at -15.0°C will not cause the solution to freeze.
To summarize, the maximum mass of fructose that can be added to 3.00 kg of pure water and still have the solution freeze at -15.0°C is zero.
To determine the maximum mass of fructose that can be added to water and still have the solution freeze at -15.0 °C, you need to use the concept of freezing point depression. The formula for freezing point depression is ΔT = Kf * m * i, where ΔT is the change in freezing point, Kf is the molal freezing point depression constant, m is the molality of the solute, and i is the van't Hoff factor.
In this case, since fructose is a molecular solid that does not ionize when dissolved in water, the van't Hoff factor (i) is equal to 1. The molal freezing point depression constant (Kf) for water is indeed 1.86 °C/m, which means that for each mole of solute per kilogram of solvent, the freezing point of water is depressed by 1.86 °C.
However, it's important to note that the Kf value is specific to the solvent being used. In this case, we are dealing with water as the solvent, so you should use the Kf value for water, which is 1.86 °C/m, rather than the Kf value for benzene.
To calculate the maximum mass of fructose that can be added, you can rearrange the freezing point depression formula to solve for the molality (m):
m = ΔT / (Kf * i)
Given that the freezing point depression (ΔT) is -15.0 °C and i is 1, you can substitute these values into the formula:
m = -15.0 °C / (1.86 °C/m * 1)
This gives you the molality of the fructose solution.
To convert this molality to mass, you need to know the molecular weight of fructose (C6H12O6), which is 180.16 g/mol. Multiply the molality by the molecular weight to get the mass of fructose per kilogram of water.
Finally, to obtain the maximum mass of fructose that can be added to 3.00 kg of water, multiply the mass per kilogram by 3.00 kg to account for the total mass of the solution.
I hope this explanation helps clarify why you should use the Kf value for water and how to calculate the maximum mass of fructose that can be added.