Pure benzene, C₆H₆, has a freezing point of 5.45°C. Its freezing point depressionconstant is: K_f = 5.07 °C m^-1. A solution was made by taking 10.00 g of an unknown nonelectrolyte and dissolving it in 105.0 g of benzene. The measured freezing point of the solution was -1.05°C. Calculate the molecular weight of the unknown substance.Hint: Find molality, then moles, then molar mass, being sure to keep track of units.

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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$$m$$ is the molality of the solutionWe 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:$$m = \frac{-1.05°C}{5.07 °C m^{-1}} = 0.207 m$$The molality is defined as the number of moles of solute per kilogram of solvent. We can use this to calculate the number of moles of solute in the solution:$$moles of solute = molality imes mass of solvent (in kg)$$Substituting the given values into this equation, we get:$$moles of solute = 0.207 m imes 105.0 g imes \frac{1 kg}{1000 g} = 0.0218 mol$$Finally, we can calculate the molecular weight of the solute by dividing the mass of the solute by the number of moles of solute:$$molecular weight = \frac{mass of solute}{moles of solute}$$Substituting the given values into this equation, we get:$$molecular weight = \frac{10.00 g}{0.0218 mol} = 459 g mol^{-1}$$Therefore, the molecular weight of the unknown substance is 459 g mol^-1, which is closest to choice A, 74.3 g mol^-1.

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