When a 1.50-g sample of glutamic acid is dissolved in 100.0 g of H₂O, the resulting solution freezes at -0.190°C. K_f for H₂O is 1.86°C/m. The molar mass of glutamic acid is A) 28.0 g/mol B) 147 g/mol C) 1.50 g/mol D) 189 g/mol E) 14.7 g/mol

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Quizplus · Accepted Answer

First, we need to calculate the molality of the solution using the freezing point depression formula: ΔTf = Kf · molality ΔTf = -0.190°C Kf = 1.86°C/m molality = ΔTf / Kf = -0.190°C / (1.86°C/m) = -0.102 mol/kg Now, we can use the molality formula to calculate the number of moles of glutamic acid in the solution: molality = moles of solute / mass of solvent in kg moles of solute = molality × mass of solvent in kg = -0.102 mol/kg × 0.1000 kg = -0.0102 mol However, we can see that we obtained a negative value for the number of moles, which does not make sense. This means that we made an error somewhere. The error is actually in the assumption that the freezing point depression is solely due to the presence of the solute (glutamic acid) in the solution. In fact, H₂O also contributes to the freezing point depression. We need to use the van't Hoff factor (i) to correct our calculation: ΔTf = Kf · molality · i H₂O is a non-electrolyte, so i = 1. Glutamic acid, on the other hand, is an amino acid that dissociates into ions in water, so i is greater than 1. The actual value of i depends on the extent of dissociation, which we do not know. However, we can assume that it is close to 1 for the sake of this problem. Let's assume i = 1, and redo the calculation: ΔTf = Kf · molality · i = 1.86°C/m × 0.102 mol/kg × 1 = 0.190°C Now, we obtain a positive value for ΔTf, which is consistent with our intuition. The corrected number of moles of glutamic acid is: molality = moles of solute / mass of solvent in kg moles of solute = molality × mass of solvent in kg = 0.102 mol/kg × 0.1000 kg = 0.0102 mol Finally, we can calculate the molar mass of glutamic acid: molar mass = mass / moles = 1.50 g / 0.0102 mol = 147 g/mol Therefore, the correct answer is B.

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