Answer:
The correct answer is 146 g/mol
Explanation:
<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
<em>Therefore, the molar mass of the compound is 146 g/mol </em>
