What is the molar mass of an aromatic hydrocarbon if 0.75 g of the compound depresses the freezing point of 123 g of benzene by 0.28°C? (K_f for benzene is 5.12°C/m.) A)43 g/mol B)150 g/mol C)110.00 g/mol 110 D)2900 g/mol E)140 g/mol

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution using the following formula:

molality = (mass of solute / molar mass of solvent) / kg of solvent

We know the mass of the solute (0.75 g) and the solvent (123 g of benzene), as well as the freezing point depression constant for benzene (5.12°C/m). We also know that the freezing point was depressed by 0.28°C.

0.28°C = Kf x molality
molality = 0.28°C / 5.12°C/m = 0.0547 m

Now, we can rearrange the formula above to solve for the molar mass of the aromatic hydrocarbon:

molar mass of solvent = (mass of solute / molality) / kg of solvent

molar mass of solvent = (0.75 g / 0.0547) / 0.123 kg = 110.00 g/mol

Therefore, the answer is choice C, 110.00 g/mol.

What Is the Molar Mass of an Aromatic Hydrocarbon If