Final answer:
The molar mass of the aromatic hydrocarbon is determined to be 111.54 g/mol by calculating the molality from the given freezing point depression, and then using the mass of the hydrocarbon and benzene to find the number of moles and ultimately the molar mass.
Step-by-step explanation:
The question asks about determining the molar mass of an aromatic hydrocarbon based on its effect on the freezing point of benzene. Using the freezing point depression formula ΔTf = Kf × m, where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent (benzene, in this case), and m is the molality of the solution, we can find the molality first. Since 0.75 g of the hydrocarbon depresses the freezing point of benzene by 0.28°C, and Kf for benzene is 5.12°C/m, the first step is to calculate the molality (m).
Using the formula:
ΔTf = Kf × m
0.28°C = 5.12°C/m × m
m = 0.28°C / 5.12°C/m
m = 0.0546875 mol/kg
Now, we convert the mass of benzene from grams to kilograms:
123 g = 0.123 kg
Using the molality (mol/kg), the mass of benzene (kg), and the mass of the hydrocarbon (g), we calculate the molar mass (M) of the aromatic hydrocarbon:
molality (m) = moles of solute / mass of solvent (kg)
0.0546875 mol/kg = moles of hydrocarbon / 0.123 kg
moles of hydrocarbon = 0.0546875 mol/kg × 0.123 kg
moles of hydrocarbon = 0.00672575 mol
Finally, molar mass (M) = mass of hydrocarbon (g) / moles of hydrocarbon
M = 0.75 g / 0.00672575 mol
M = 111.54 g/mol
Therefore, the molar mass of the aromatic hydrocarbon is 111.54 g/mol.