What is the freezing point of an aqueous solution of 10.0 g of glucose (C6H12O6) in 50.0g H2O? (kf=-1.86 C/m)
what is m*Kf?
Ans: it is -1.86*10/(180*.050)
check that mole mass for the sugar.
Oh, you're really putting me on the spot here. You want to know the freezing point of a solution? Well, let me think... Did you know that jokes have a lower freezing point than regular facts? Yep, it's true! Okay, okay, I'll give you the answer. So, the freezing point depression is calculated using the formula 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 solution, and i is the Van't Hoff factor. Now, let me crunch some numbers here... calculating... calculating... Ah-ha! The molality of the solution can be found by dividing the moles of solute by the mass of the solvent. The moles of glucose can be calculated using its molar mass and the mass given. And the mass of the solvent is, well, the mass given. Once you've got the molality, you can plug everything into the ΔT formula and solve for the change in freezing point. So the freezing point of the solution would be the freezing point of pure water minus the change in freezing point. Phew! That was quite a journey, wasn't it?
To find the freezing point of the solution, we need to use the formula:
ΔTf = -(kf * m)
where ΔTf is the change in freezing point, kf is the freezing point depression constant, and m is the molality of the solution.
First, let's find the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
To find moles of solute (glucose):
moles = mass / molar mass
The molar mass of glucose (C6H12O6) is:
6*(12.01 g/mol) + 12*(1.01 g/mol) + 6*(16.00 g/mol) = 180.18 g/mol
Now, let's calculate the moles of glucose:
moles of glucose = 10.0 g / 180.18 g/mol
Next, let's convert the mass of the solvent (H2O) to kg:
mass of H2O = 50.0 g / 1000 = 0.0500 kg
Finally, we can find the molality (m):
m = moles of solute / mass of solvent
m = (10.0 g / 180.18 g/mol) / 0.0500 kg
Now, we can substitute the values in the formula to find the change in freezing point (ΔTf):
ΔTf = -(kf * m)
ΔTf = -(1.86 C/m * [(10.0 g / 180.18 g/mol) / 0.0500 kg])
Calculating this expression will give us the change in freezing point. To find the freezing point of the solution, we need to subtract this value from the freezing point of pure water (0°C or 273.15 K).
To find the freezing point of the aqueous solution, we need to use the concept of freezing point depression. This concept states that the freezing point of a solution is lower than that of the pure solvent. The amount by which the freezing point is lowered depends on the molality of the solute.
First, we need to calculate the molality (m) of the glucose solution. Molality is defined as the number of moles of solute per kilogram of solvent.
Step 1: Calculate the number of moles of glucose (C6H12O6)
We have 10.0 g of glucose. The molar mass of glucose (C6H12O6) is approximately 180 g/mol.
Number of moles = mass / molar mass
Number of moles of glucose = 10.0 g / 180 g/mol = 0.0556 mol
Step 2: Calculate the mass of water (H2O)
We have 50.0 g of water.
Step 3: Calculate the molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent.
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of water in kg = 50.0 g / 1000 g/kg = 0.05 kg
Molality (m) = 0.0556 mol / 0.05 kg = 1.11 mol/kg
Step 4: Calculate the freezing point depression (∆Tf)
The equation for freezing point depression is given by ∆Tf = Kf × m
where ∆Tf is the change in freezing point, Kf is the cryoscopic constant for the solvent, and m is the molality of the solution.
Given: Kf = -1.86 °C/m (given in the problem)
∆Tf = -1.86 °C/m × 1.11 mol/kg = -2.0646 °C
Step 5: Calculate the freezing point of the solution
The freezing point of the pure solvent (water) is 0 °C.
Freezing point of solution = freezing point of pure solvent + ∆Tf
Freezing point of solution = 0 °C + (-2.0646 °C) = -2.0646 °C
Therefore, the freezing point of the aqueous solution of 10.0 g of glucose in 50.0 g of water is approximately -2.06 °C.