A 6.06-gram sample of a compound is dissolved in 250.grams of benzene.The freezing point of this solution is 1.02°C below that of pure benzene.What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m. )Ignore significant figures for this problem. A)30.4 g/mol B)122 g/mol C)243 g/mol D)4.83 g/mol E)60.8 g/mol

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Quizplus · Accepted Answer

The freezing point depression ($\Delta T_f$) is given as 1.02°C. The molal freezing point depression constant ($K_f$) for benzene is 5.12°C/m. Using the formula $\Delta T_f = K_f \times m$, where $m$ is the molality of the solution, we can find the molality: $m = \frac{\Delta T_f}{K_f} = \frac{1.02°C}{5.12°C/m} = 0.1992 m$. Molality is defined as moles of solute per kilogram of solvent, so $0.1992 = \frac{\text{moles of solute}}{0.250 \text{ kg}}$. Solving for moles of solute gives $0.0498 \text{ moles}$. The molar mass ($M$) of the compound is then calculated using the mass of the sample divided by the moles of solute: $M = \frac{6.06 \text{ g}}{0.0498 \text{ moles}} = 121.69 \text{ g/mol}$, which rounds to 122 g/mol.

A 606-Gram Sample of a Compound Is Dissolved in 250