Formula: Tf = Kf * m Where Tf is the depression in the freezingpoint Kf is the m

Question

Formula:             Tf = Kf * m Where Tf is the depression in the freezingpoint             Kf is the molal depression in freezing point constant             m is the molality Data:                                      Mass = 550.0 g Molar mass of ethylene glycol = 62.068 g/mol Number of moles of ethylene glycol = 550.0 g / 62.068 g /mol                                                       =8.86 moles                             Mass of solvent = 3.245 kg                                        Molality = 8.86 moles / 3.245 kg                                                     = 2.73 m                                           Tf = 1.86 0 C / m * 2.73 m                                                      =5.1 0 C                                           Tsoln = 0 0C - 5.1 0 C                                                     =- 5.1 0 C Formula:                  Tb= Kb * m Where Tb is the elevation in thefreezing point             Kf is the molal elevation in point constant             m is the molality Data:                                      Mass = 550.0 g Molar mass of ethylene glycol = 62.068 g/mol Number of moles of ethylene glycol = 550.0 g / 62.068 g /mol                                                       =8.86 moles                             Mass of solvent = 3.245 kg                                        Molality = 8.86 moles / 3.245 kg                                                     = 2.73 m                                           Tb = 0.52 0 C / m * 2.73 m                                                      = 1.410 C                                           Tsoln = 100 0C  +1.41 0C                                                     = 101.40 C Formula:                  Tb= Kb * m Where Tb is the elevation in thefreezing point             Kf is the molal elevation in point constant             m is the molality Data:                                      Mass = 550.0 g Molar mass of ethylene glycol = 62.068 g/mol Number of moles of ethylene glycol = 550.0 g / 62.068 g /mol                                                       =8.86 moles                             Mass of solvent = 3.245 kg                                        Molality = 8.86 moles / 3.245 kg                                                     = 2.73 m                                           Tb = 0.52 0 C / m * 2.73 m                                                      = 1.410 C                                           Tsoln = 100 0C  +1.41 0C                                                     = 101.40 C Where Tb is the elevation in thefreezing point             Kf is the molal elevation in point constant             m is the molality Data:                                      Mass = 550.0 g Molar mass of ethylene glycol = 62.068 g/mol Number of moles of ethylene glycol = 550.0 g / 62.068 g /mol                                                       =8.86 moles                             Mass of solvent = 3.245 kg                                        Molality = 8.86 moles / 3.245 kg                                                     = 2.73 m                                           Tb = 0.52 0 C / m * 2.73 m                                                      = 1.410 C                                           Tsoln = 100 0C  +1.41 0C                                                     = 101.40 C

Explanation / Answer

One error sticks out to me: molality = (mols solute) / (kg solution) You are dividing the mols solute by kg solvent. You need toinclude the mass of the ethylene glycol in the total mass of thesolution so: 8.86mol / (3.245kg + .550kg) = 8.86mol / 3.795kg = molality of the solution Everything else looks good.

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