0.15 grams of benzene (C6H6) is dissolved into 4.7 grams of melted biphenyl. The freezing point of the mixture was determined to be 4.7 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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To determine the Kf (molal freezing point depression constant) of biphenyl, we can use the formula:

ΔTf = Kf * m * i

Where:
ΔTf is the freezing point depression of the solution,
Kf is the molal freezing point depression constant,
m is the molality of the solute (benzene in this case),
i is the van't Hoff factor (the number of particles the solute dissociates into).

Given that the freezing point depression of the mixture is 4.7 degrees, and assuming benzene does not dissociate, i can be taken as 1.

First, we need to calculate the molality (m) of benzene in the solution.

Molar mass of benzene (C6H6) = 6*12.01 + 6*1.01 = 78.11 g/mol
Number of moles of benzene = mass / molar mass = 0.15 g / 78.11 g/mol ≈ 0.00192 mol

Molar mass of biphenyl = 2*12.01 + 10*1.01 = 154.20 g/mol
Number of moles of biphenyl = mass / molar mass = 4.7 g / 154.20 g/mol ≈ 0.0305 mol

Molality (m) of benzene = moles of solute / mass of solvent (in kg)
Since biphenyl is the solvent: m = 0.00192 mol / 4.7 g ≈ 0.000409 mol/kg

Now, we can plug these values into the freezing point depression equation:

4.7 degrees = Kf * 0.000409 mol/kg * 1

Solving for Kf:

Kf = 4.7 degrees / (0.000409 mol/kg)
Kf ≈ 11,498 degrees * kg / mol

Therefore, the Kf of biphenyl, based on the experimental results, is approximately 11,498 degrees * kg / mol.