The vapor pressure of pure water at 25 degree C is 23.8 torr. Determine the vapo

Question

The vapor pressure of pure water at 25 degree C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 degree C above a solution prepared by dissolving 25 g of urea (a nonvolatile, non-electrolyte, MW = 60.0 g/mol) in 75 g of water. A. 2.9 B. 0.42 C. 22 D. 27 E. 0.91 The freezing point of ethanol (C_2H_5OH) is -114.6 degree C. The molal freezing point depression constant for ethanol is 2.00 degree C/m. What is the freezing point (degree C) of a solution prepared by dissolving 50.0 g of glycerin (C_3H_8O_3, a nonelectrolyte) in 200.0 g of ethanol? A. -115.0 B. -5.43 C. -132.3 D. -120.0 E.-114.6

Explanation / Answer

(2)

Moles of water in solution = 75/18 = 4.17

Moles of urea = 25/60 = 0.417

So, mole fraction of water, Xw = 4.17/(4.17+0.417) = 0.91

So, Using Raoult's law, vapor pressure of water, Pw = Xw*PH2O

Given: PH2O = 23.8

So, Pw = 0.91*23.8 = = 21.66 = 22 torr

(3)

Moles of glycerin = 50/MW = 50/92 = 0.543

So, molality of solution, m = Moles of solute/Mass of solvent in kgs = 0.543/0.2 = 2.715 molal

Using the equation:

Tf-Ti = -Kf*m

Tf = Ti - Kf*m = -114.6 - 2*2.715 = -120.03 = -1200C

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