0.24 grams of benzene (C6H6) is dissolved into 4.9 grams of melted biphenyl. The freezing point of the mixture was determined to be 4.4 degrees below the freezing point of pure biphenyl. Based on these results, what is the Kf of biphenyl?
(Use the standard units in your calculation, oC/m, but do not include the units in your answer. Also note these are experimental values so they may not equal the literature value for Kf).
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mol benzene = grams/molar mass benzene. Substitute and solve for mol benzene.
molality = mols benzene/kg biphenyl.
Substitute and solve for molality.
delta T = Kf*m
You know delta T and you know m, solve for Kf.
delta T, of course, is 4.4.
Post your work if you get stuck.
To calculate the Kf (molal freezing point depression constant) of biphenyl, we can use the formula:
∆T = Kf * m
Where:
- ∆T is the change in freezing point of the solution
- Kf is the molal freezing point depression constant
- m is the molality of the solute
From the information given, we know that the freezing point of the mixture was determined to be 4.4 degrees below the freezing point of pure biphenyl. This means ∆T = -4.4°C.
We need to calculate the molality (m) of benzene in the solution. Molality is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the moles of benzene:
moles of benzene = mass of benzene / molar mass of benzene
The molar mass of benzene (C6H6) can be found by adding up the atomic masses of its constituent elements:
molar mass of benzene = (6 * atomic mass of carbon) + (6 * atomic mass of hydrogen)
The atomic masses can be found from the periodic table.
Next, we calculate the molality:
molality = moles of benzene / mass of biphenyl (in kg)
Finally, we can rearrange the formula to solve for Kf:
Kf = ∆T / m
Substituting the given values, we should be able to solve for Kf.