plusminus Boiling-Point Elevation and Freezing-Point Depression for Solutions in
ID: 1023182 • Letter: P
Question
plusminus Boiling-Point Elevation and Freezing-Point Depression for Solutions in Water 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 pant 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, Delta T_b, can be calculated using the equation Delta T_b = K_b middot m in which m is the molality of the solution and K_b is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, Delta T_f, can be calculated in a similar manner using Delta T_f =- K_f middot m in which m is the molality of the solution and K_f is the molal freezing-point-depression constant for the solvent. Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C_2 H_6 O_2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze. What is the freezing point of radiator fluid that is 50% antifreeze by mass? K_f for water is 1.86 degree C/m. Express your answer to three significant figures and include the appropriate units. What is the boiling point of radiator fluid that is 50% antifreeze by mass? K_b for water is 0.512 degree C/m. Express your answer to one decimal place and include the appropriate units.Explanation / Answer
1]
Kf for water is 1.86 degree Celsius C/m
It Molar mass C2H6O2 = 62.0g/mol
1000g = 1000/62 = 16.12 mol per kg water
solution is 16.12m
Freezing point depression = 1.86*16.13 = 29. 98 °C
The solution will freeze at = -30 C
2]
50% = 50 g C2H6O2 per 50 g H2O
Molar mass,
50 g C2H6O2 = 62.068 g/mol = 50/ 62.068 = 0.8056 moles
0.0056 moles / 0.050 kilograms H2O = 16.11 mole solution
16.11 mole 0.512C/m.? = 8.25 C rise in temp
= 108.2 C
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