Pure benzene freezes at 5.45°C. A solution containing 724g of C2Cl4H4 in 115.3 g
of benzene was observed to freeze at 3.55°C. From these data, calculate the molal
freezing point depression constant of benzene.
Perhaps no one has answered because C2Cl4H4 does not exist. Check it out. That would mean C has 5 bonds and that's a no no.
To calculate the molal freezing point depression constant of benzene (Kf), we can use the formula:
ΔTf = Kf * m
Where:
ΔTf = freezing point depression (in °C)
Kf = molal freezing point depression constant (in °C * kg/mol)
m = molality (in mol/kg)
First, we need to calculate the molality (m).
1. Calculate the number of moles of C2Cl4H4:
We know that the molar mass of C2Cl4H4 is 98 g/mol.
Number of moles = mass / molar mass
Number of moles of C2Cl4H4 = 724 g / 98 g/mol
Number of moles of C2Cl4H4 = 7.39 mol
2. Calculate the molality (m):
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of solvent = 115.3 g = 0.1153 kg
Molality (m) = 7.39 mol / 0.1153 kg
Molality (m) ≈ 64.12 mol/kg
The molality (m) is approximately 64.12 mol/kg.
Now, we can use the freezing point depression to calculate the molal freezing point depression constant (Kf).
ΔTf = 5.45°C - 3.55°C
ΔTf = 1.9°C
Plugging in the values into the formula:
1.9°C = Kf * 64.12 mol/kg
Solving for Kf:
Kf ≈ 1.9°C / 64.12 mol/kg
Kf = 0.0296 °C * kg/mol
The molal freezing point depression constant of benzene (Kf) is approximately 0.0296 °C * kg/mol.
To solve this question, you need to use the equation for freezing point depression:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point
Kf is the molal freezing point depression constant
m is the molality of the solution
First, we need to calculate the change in freezing point, which is the difference between the freezing point of a pure solvent (benzene) and the observed freezing point of the solution.
ΔTf = 5.45°C - 3.55°C
= 1.90°C
Next, we need to calculate the molality (m) of the solution. Molality is the number of moles of solute (C2Cl4H4) divided by the mass of the solvent (benzene) in kilograms.
First, let's calculate the number of moles of C2Cl4H4:
molar mass of C2Cl4H4 = (12.01 g/mol * 2) + (1.01 g/mol * 4) + (35.45 g/mol * 4) + (1.01 g/mol * 4)
= 112.53 g/mol
moles of C2Cl4H4 = 724 g / 112.53 g/mol
≈ 6.439 mol
Next, let's calculate the mass of benzene in kilograms:
mass of benzene = 115.3 g / 1000
= 0.1153 kg
Now, let's calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
= 6.439 mol / 0.1153 kg
≈ 55.9 mol/kg
Finally, we can substitute the values into the equation:
1.90°C = Kf * 55.9 mol/kg
Now rearrange the equation to solve for Kf:
Kf = ΔTf / m
= 1.90°C / 55.9 mol/kg
Now you can calculate Kf.