Question

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.

Answer 1

18.700 g

Step-by-step explanation:

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

`P_(vs) =P_(w) *x_(w)`

Where
`P_(vs)` is the vapor pressure of the solution,
`P_(w)` is the vapor pressure of the pure water, and
`x_(w)` 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:

`23.22mmHg=23.76mmHg*x_(w)\\ x_(w)=(23.22mmHg)/(23.76mmHg)=0.977`

So, the molar fraction of the urea should be:
`x_(urea)=1-x_(w)=0.023`

Then, calculate the average molecular weight:

`M=x_(w)*MW_(w)+x_(urea)*MW_(urea)\\ M=0.977*18.02+0.023*60.10=18.989`

The molar fraction of urea is:

`0.023=(X urea mol)/(S solution moles)=(x urea grams)/(238.2+x (solution grams))*(1 urea mol)/(60.10 g)*(18.989 solution grams)/(1 solution mol)`

Solving for x,

`x=18.700g`