A 5.96-g sample of a compound is dissolved in 222.0 g of benzene. The freezing point of this solution is 1.06°C below that of pure benzene. What is the molar mass of this compound? (Note: K_f for benzene = 5.12°C/m.) A) 1.30* 10² g/mol B) 2.85 *10¹g/mol C) 1.80 * 10² g/mol D) 6.39 * 10³ g/mol E) 5.56 *10² g/mol

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Quizplus · Accepted Answer

The freezing point depression is given by the equation:$$\Delta T_f = K_f 	imes m$$where $\Delta T_f$ is the freezing point depression, $K_f$ is the freezing point depression constant, and $m$ is the molality of the solution.We can rearrange this equation to solve for the molality:$$m = \frac{\Delta T_f}{K_f}$$Substituting the given values into this equation, we get:$$m = \frac{1.06°C}{5.12°C/m} = 0.207 m$$The molality is defined as the number of moles of solute per kilogram of solvent. We can use this to calculate the molar mass of the compound:$$M = \frac{m 	imes MW_{solvent}}{mass_{solute}}$$where $M$ is the molar mass of the compound, $MW_{solvent}$ is the molecular weight of the solvent, and $mass_{solute}$ is the mass of the solute.Substituting the given values into this equation, we get:$$M = \frac{0.207 m 	imes 78.11 g/mol}{5.96 g} = 2.85 	imes 10^2 g/mol$$Therefore, the molar mass of the compound is 2.85 x 10^2 g/mol, which corresponds to choice B.

A 596-G Sample of a Compound Is Dissolved in 222