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 car 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 balling 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 F_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_2H_6O_2. The radiator fluid used in most cars in a half and half mixture of water and antifreeze. What is the freezing point of radiator fluid that is 60% antifreeze by mass? K_f for water is 1.86 degree C/m. What is the boiling point of radiator fluid that is 50% antifreeze by mass? K_b for water is 0.512 degree C/m.

Explanation / Answer

Freezing point depression = i*kf*m, i= Van;t Hoff factor =1 for ethylene glycol

kf= 1.86 deg.c/m, m= molality = moles of solute/ kg of solvent

basis : 100 gm of antifreeze mixture. It contains 50 gm ethylene glycol and 50 gm water ( 50/1000 kg water =0.05 kg water). moles of ethylene glycol= mass/molar mass = 50/62 =0.81

molality, m= 0.81/0.05 =16.2

Depression in freezing point = 1*1.86*16.2 =30.13, Freezing point =0(water)-30.13= -30.13 deg.c

2. Boiling point elevation = i*kb*m= 1*0.512*16.2= 8.3 deg.c

elevation in boiling point =100+8.3= 108.3 deg.c

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