What is the boiling-point change for a solution containing 0.308 mol of naphthalene (a nonvolatile, nonionizing compound) in 266.0 g of liquid benzene? (K_b = 2.53°C/m for benzene) A) 2.18°C B) 7.14°C C) 0.458°C D) 0.207°C E) 2.93°C

Question

Quizplus · Accepted Answer

We can use the equation ΔTb = Kb · m · i, where ΔTb is the boiling point elevation, Kb is the molal boiling-point elevation constant, m is the molality of the solution (moles of solute per kilogram of solvent), and i is the van 't Hoff factor, which is 1 for nonionizing solutes like naphthalene. First, we need to calculate the molality of the solution using the formula

molality = moles of solute / mass of solvent in kg

moles of naphthalene = 0.308 mol
mass of benzene = 266.0 g = 0.2660 kg

molality = 0.308 mol / 0.2660 kg ≈ 1.16 mol/kg

Now we can plug in the values to find ΔTb:

ΔTb = (2.53°C/m) · (1.16 mol/kg) · 1
ΔTb ≈ 2.93°C

Therefore, the boiling-point change for the solution is 2.93°C, which corresponds to answer choice (E).

What Is the Boiling-Point Change for a Solution Containing 0