"Cowboy coffee" is prepared by boiling coarse coffee grounds in water. If 12.5 g sucrose (C12H22O11) is added to 225 g cowboy coffee as it is being prepared, by how much is the vapor pressure above the solution reduced? Assume the solvent, the cowboy coffee, has the same properties as water. The vapor pressure of the solvent at 100 °C is 760.0 Torr.
- hcbiochemLv 73 months agoFavorite Answer
Moles sucrose = 12.5 g / 342.3 g/mol = 0.0365 mol
moles H2O = 225 g / 18.01 g/mol =12.5 mol water
mole fraction water = (12.5 / (12.5+0.0365) = 0.997
Vapor pressure solution = 760.0 torr X 0.997 = 757.8 torr
So, the vapor pressure is reduced by 2.2 torr
- 3 months ago