The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 8.50 g of naphthalene (C10H8) in 425 g of benzene (Kf of benzene = 4.90°C/m)?
I worked this problem for someone just a couple of days ago.
delta T = kf m
kf = 4.90
m = molality = mols/kg. Convert 8.50 g naphthalene to mols and divide by kg of benzene (0.425).
This will give you delta T. Subtract from 5.5 to give you the new freezing point. Check my work. Check my thinking.
To find the freezing point of the solution, we can use the formula:
ΔT = Kf * m
In this formula, ΔT represents the change in temperature (in °C), Kf is the freezing point depression constant (in °C/m) for the solvent (benzene in this case), and m is the molality of the solution.
Given that Kf for benzene is 4.90 °C/m, we can plug in the values into the formula:
ΔT = 4.90 °C/m * m
Now, we need to calculate the molality of the solution. Molality is defined as the amount of solute (in moles) divided by the mass of the solvent (in kg).
First, let's calculate the number of moles of naphthalene (C10H8). We have 8.50 g of naphthalene. To convert grams to moles, we need to know the molar mass of naphthalene.
The molar mass of naphthalene (C10H8) is the sum of the individual atomic masses of carbon (C) and hydrogen (H). It can be calculated as follows:
Molar mass of C = 12.01 g/mol * 10 = 120.1 g/mol
Molar mass of H = 1.01 g/mol * 8 = 8.08 g/mol
Total molar mass of naphthalene = 120.1 g/mol + 8.08 g/mol = 128.18 g/mol
So, 8.50 g of naphthalene is equal to (8.50 g) / (128.18 g/mol) = 0.0662 mol
Next, we need to calculate the mass of benzene in kg. We have 425 g of benzene, which is equal to 0.425 kg.
Now, let's calculate the molality of the solution:
m = (moles of solute) / (kg of solvent)
m = 0.0662 mol / 0.425 kg ≈ 0.1554 mol/kg
Now that we have the molality (m), we can calculate the change in temperature (ΔT):
ΔT = 4.90 °C/m * 0.1554 mol/kg ≈ 0.76 °C
Lastly, to find the new freezing point of the solution, we subtract the change in temperature (ΔT) from the freezing point of benzene:
Freezing point of solution = 5.5 °C - 0.76 °C ≈ 4.74 °C
Therefore, the freezing point of the solution of 8.50 g of naphthalene in 425 g of benzene is approximately 4.74 °C.