What is the boiling-point change for a solution containing 0.251 mol of naphthalene (a nonvolatile,nonionizing compound)in 250.g of liquid benzene? (K_bp = 2.53°C/m for benzene) A) 2.54 °C B) 10.08 °C C) 0.159 °C D) 2.52 °C E) 0.635 °C

Question

Quizplus · Accepted Answer

First, we need to use the equation:
ΔT = iKfm
where ΔT is the boiling-point elevation, i is the van't Hoff factor (equal to 1 for nonionizing compounds like naphthalene), Kf is the freezing-point depression constant (given as 2.53°C/m for benzene), and m is the molality of the solution.
We can calculate the molality of the solution using:
molality = moles of solute / mass of solvent (in kg)
moles of naphthalene = 0.251 mol
mass of benzene = 250 g = 0.25 kg
molality = 0.251 mol / 0.25 kg = 1.004 mol/kg
Now we can plug in the values to find the boiling-point elevation:
ΔT = 1 x 2.53°C/m x 1.004 mol/kg = 2.54°C
Therefore, the boiling-point elevation for the solution is 2.54°C, which is closest to choice A (2.54 °C).

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