A solution containing 20.0 g of an unknown liquid and 110.0 g water has a freezing point of -1.32 °C. Given K_f = 1.86 °C/m for water, the molar mass of the unknown liquid is ________ g/mol. A)256 B)69.0 C)333 D)619 E)78.1

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The correct answer is A.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 for the solvent$$m$$ is the molality of the solutionWe can rearrange this equation to solve for the molality of the solution:$$m = \frac{\Delta T_f}{K_f}$$We are given the following information:$$\Delta T_f = -1.32 °C$$$$K_f = 1.86 °C/m$$We can substitute these values into the equation to solve for the molality of the solution:$$m = \frac{-1.32 °C}{1.86 °C/m} = 0.71 m$$The molality of the solution is defined as the number of moles of solute per kilogram of solvent. We are given that the solution contains 20.0 g of an unknown liquid and 110.0 g of water. We can use this information to calculate the number of moles of solute in the solution:$$moles \ of \ solute = 20.0 g 	imes \frac{1 mol}{molar \ mass}$$We can substitute the molality of the solution into this equation to solve for the molar mass of the unknown liquid:$$molar \ mass = \frac{20.0 g}{0.71 m} = 28.2 g/mol$$Therefore, the molar mass of the unknown liquid is 28.2 g/mol.

A Solution Containing 20