Vapour Pressure of a Solution of a Non-Volatile Solute.?

Question

A solution is made by dissolving 49.0g of urea (H2NCONH2, a non-volatile non-electrolyte) in 256g of water.
Calculate the vapour pressure of this solution at 65°C.
(note the vapour pressure of pure water at 65°C is 187.5 torr.)
okay so I can use Raoults Law but i am confused.
P solution= X solvent*P solvent
so how do i arrange it i know P solvent is 187.5 help please!!

Answer 1

This answer will be in torr if you use 187.5, for good measure I would convert this back to Pascals.



MM of Urea = 60g/mole

# moles of urea = .82 moles

# moles of water = 256/16 = 16 moles



16/16.82 = X of H20 = 0.95 (95% of moles are water)



so answer will be 95% of pure vapor pressure is solution vapor pressure --> 0.95*187.5 = 178 torr or 23.7 kPa



This solvent vapor pressure will be relatively accurate, Raoult's law applies well for high solvent concentrations



@ Additional details :



No, look at the calculation. The mole fraction of urea is 5%. Let's start over.



P(solution total/AKA ur answer) = P(pure water pressure) * X (mole fraction of water) + P(pure urea pressure) * X(mole fraction of urea)





Pure Urea Pressure = 0

so equation is just the water component. You multiply the mole fraction of WATER by the pure water solvent pressure. This lowers the total solution pressure from that of pure water. Read my answer again carefully. You will not succeed if you don't read carefully.

Answer 2

first find mols of water and moles of urea

ie. 256g/18.02g/mol and 49g/60.06g/mol

then find mole fraction of solvent (ie. water)

---->>>> mols water/(mols water+mols urea)

then, P of solution = mole fraction * P solvent (this is given= 187.5)

Answer 3

moles urea = 49.0 g/ 60.06 g/mol=0.816

moles water = 256 g/ 18.02 g/mol=14.2

mole fraction water = 14.2 / 14.2 + 0.816=0.946

p = 0.946 x 187.5 = 177.3 torr