A solution is prepared by dissolving 35.0 g glycerin (C_3H_8O_3), (MM = 92.09 g/

Question

A solution is prepared by dissolving 35.0 g glycerin (C_3H_8O_3), (MM = 92.09 g/mol) in 125 g water (MM = 18.02 g/mol). Assume that glycerin is a nonvolatile solute. What is the vapor pressure of this solution at 25.0 degree C? The vapor pressure of pure water is 23.76 mm Hg at 25.0 degree C. Windshield wiper fluid is approximately a 40.0% by mass methanol (CH_3OH, MM = 32.042 g/mol) in water solution. The methanol is added to depress the freezing point of water to remove ice 011 the windshield. Calculate the freezing point of the windshield wiper solution, given that pure water has a normal freezing point of 0.00 degree C and a molal freezing-point depression constant (K_f) is 1.86 degree C/m.

Explanation / Answer

1) As per Raoult's law, fraction change in vapor pressure = mole fraction of the non-volatile solute

Now, moles of 35 g of glycerin = mass/molar mass = 35/92.09 = 0.38

moles of 125 g of water = mass/molar mass = 125/18.02 = 6.94

Thus, mole fraction of solute = moles of glycerin/(moles of glycerin + moles of water) = 0.052

Thus, as per Raoult's law, (23.76 - P)/23.76 = 0.052

or, P = vapor pressure of the solution = 22.53 mm of Hg

2) Let there be 100 g of the solution

Therefore mass of methanol = 40 g

moles of methanol = mass/molar mass= 40/32.042 = 1.25

molality of the solution = moles of methanol/mass of water in kg = 1.25/0.06 = 20.81 m

Thus, depression in F.P of the solution = Kf*m = 1.86*20.81 = 38.7

Thus, F.P of the solution = -38.7 0C

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