The boiling point and freezing points of a solution differs from those of the pu

Question

The boiling point and freezing points of a solution differs from those of the pure liquid. This can be explained in terms of vapor pressure. Since the vapor pressure of the solvent above the solution is lower, a higher temperature is needed to reach a point where the vapor pressure of the liquid meets the required 1 atm, and the boiling point is elevated. The lower vapor pressure changes the entire phase diagram for the solvent, and the resulting change pushes the triple point of the solution to a lower temperature value. The solid-liquid phase equilibria curve is related to the location of the triple point, and the freezing point is also lower.

The change in the boiling point for a solution containing a molecular solute, Tb, can be calculated using the equation

Tb=Kbm

in which m is the molality of the solution and Kb is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, Tf, can be calculated in a similar manner using

Tf=Kfm

in which m is the molality of the solution and Kf is the molal freezing-point-depression constant for the solvent.

Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C2H6O2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.

Part A

What is the freezing point of radiator fluid that is 50% antifreeze by mass?
Kf for water is 1.86 C/m.

Part B

What is the boiling point of radiator fluid that is 50% antifreeze by mass?
Kb for water is 0.512 C/m.

Express your answer to one decimal place and include the appropriate units.

Explanation / Answer

only respondere Part A, if you want the answer to Part B ajunta latter a new question, data are lacking to determine the molality, so let on standard data that possess 500 g of water and 500 g of ethylene glycol, we molality that:
number of moles of solute = 500g / 62g / mole = 8.06 moles
molality = 8.065 mol / 0.5 kg = 16.13 mol / kg

Tf = Kfm
Tf = 1.86 C / m16.13 mol / kg = 30.0018 mol. ° C / kg.m

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