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.

Express your answer to three significant figures and include the appropriate units.

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

C2H6O2

50%-50% by mass

we need

dT = K*m

m = molality (mol of solute/kg of solvent)

Assume a basis of 100 kg

50 kg are C2H6O2

50 kg are water

MW of C2H6O2 = 62 g

calculate moles of C2H6O2

mol = mass/MW = 50000/62 = 806.45 mol of C2H6O2

calculate molality

m = mol/kg = 806.45 / 50 kg of water = 16.13 m

Now

dT = K*m

dT = ( 1.86 C/m) * 16.13 m = 30°C

T = 0°C

Tantifreeze = T-dT = 0-30°C = -30.0°C

b)

for radiator fluid

Kb = 0.512°C/m

50% of antifreeze and 50% of water

Same properties as before so

m = 16.13 mol/kg

but this one is for boiling point so

dT = K*m

dT = 0.512°C/m * 16.13 = 8.258°C

Tnew = T+dT = 8.25°C+100 = 108.3°C

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