Question

VAPOR PRESSURE PROB!
Sucrose is a non volatile, non-ionizing solute in water.
Determine the vapor pressure lowering, at 27 degrees Celsius, of a solution of 75.0 grams sucrose, C12H22O11 dissolved in 180 grams of water.
Note: the vapor pressure of water at 27 degrees Celsius is 26.7 torr.

I've done this:
Psolvent = XsolventPºsolvent
75g sucrose / 342g/mole = 0.219moles sucrose
180g water / 18g/mole = 10moles water
mole fraction solvent= 0.978
Psolvent = (0.978) (26.7torr) = 26.13torr
What do I have to do next?! The answer is .571 torr! Any one able to help? Please and thank you!

Answer 1

P0 - P/P0= n2/n1
where Po is = 26.7 torr
find Po-P =delta P=?
n2= 0.219 moles
n1= 10moles

Answer 2

did u get it now LOL :)

Answer 3

Po-P = delta P=Po * n2/n1 = 26.1*0.219 / 10 = 0.571 torr :)