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Finding the MP & BP of solution hwne mass of solute and volume of solution is given

Submitted by nicolegurascio on Sat, 09/10/2011 - 21:48

Hello -- I'm having trouble determining how to solve the following question. Can anyone help?

A solution of urea, NH2C(O)NH2, was prepared by dissolving 35.73 g of urea in 127 mL of acetic acide. What are the Melting and boilining points of the solution?


Thank you

Somewhere in your textbook there is a chapter on solutions, mentioning freezing point depression and boiling point elevation. It will probably have one or more tables, and you will find the constants you need.

For example, in the fourth edition of the Brady & Senese Chemistry textbook, page 624 has a table that lists (for acetic acid) boiling point = 118.3°C, melting point = 16.6°C, Kb = 3.07°C/m, and Kf = 3.57°C/m.

It will explain about colligative properties, and molality, and tell you how to use molality (m) to calculate changes in boiling point (ΔTb) and freezing point (ΔTf) as you add a solute to a solvent.

The changes are calculated as ΔTb = Kbm and ΔTf = Kfm, with m = moles of solute/kg of solvent.

If you are facing that problem, you probably know very well how to figure outthe number of moles of urea in 35.75 g of urea.

You will need to figure out how many kilograms of acetic acid are in 127 mL of acetic acid.  for that you will need the density of acetic acid. That value may be harder to found. It may or may not be in atable on your textbook. You could find it on a label of a bottle of acetic acid in the lab. I found a density of 1.0498 g/mL for 100% acetic acid at 20°C, and I would use 1.050 g/mL to convert

(125 mL)(1.050 g/mL)(1 kg/1000g) = 0.131 kg

I think I got it. Would you mind confirming if I'm getting the correct #'s. ΔTb = Kbm ΔTb = (3.07°C)(44.8 m) ΔTb = 137.536 So BP would be 137.5°C ΔTf = Kfm ΔTf = (3.57°C)(44.8 m) ΔTf = 159.94 So MP would be 160°C