The vapor pressure of water is 23.76 mm hg at 25°c. how many grams of urea, ch4n2o, a nonvolatile, nonelectrolyte (mw = 60.10 g/mol), must be added to 238.2 grams of water to reduce the vapor pressure to 23.22 mm hg ? water = h2o = 18.02 g/mol.

18.700 g

Explanation:

As the urea is a nonvolatile and nonelectrolyte solute, it will reduce the vapor pressure of the solution according to:

Where is the vapor pressure of the solution, is the vapor pressure of the pure water, and is the molar fraction of water. This equation applies just for that kind of solutes and at low pressures (23.76 mmHg is a low pressure).

From the equation above lets calculate the water molar fraction:

So, the molar fraction of the urea should be:

Then, calculate the average molecular weight:

The molar fraction of urea is:

Solving for x,

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .