What is the freezing point of a solution containing 2.80 grams benzene (molar mass = 78.11 g/mol)dissolved in 43.0 grams paradichlorobenzene (molar mass = 147.0 g/mol)? The freezing point of pure paradichlorobenzene is 53.0 $\circ$ C and the freezing point depression constant,K_fp,is -7.10 $\circ$ C/m. A) 46.7 $\circ$ C B) 47.1 $\circ$ C C) 58.9 $\circ$ C D) 52.6 $\circ$ C E) 58 .9 $\circ$ C

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:

Moles of benzene = 2.80 g / 78.11 g/mol = 0.0359 mol
Moles of paradichlorobenzene = 43.0 g / 147.0 g/mol = 0.2925 mol

Molality = moles of solute / mass of solvent in kg
Molality = 0.0359 mol / 0.0430 kg = 0.834 mol/kg

Next, we can use the freezing point depression equation to calculate the change in freezing point (ΔTf):

ΔTf = Kf x molality

ΔTf = -7.10  ℃/m x 0.834 mol/kg = -5.91 ℃

Finally, we can calculate the freezing point of the solution:

Freezing point = Freezing point of pure solvent + ΔTf
Freezing point = 53.0 ℉ - (5.91 ℃ x 1 ℉/1.8 ℃) = 47.1 ℉

Therefore, the freezing point of the solution is 47.1 ℉, which corresponds to choice (B).

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