Question
Why would [23.8-(.0588)(23.8)] be the proper setup to determine the vapor pressure of a solution at 25 Celsius that has 45g of C6H12O6, glucose (MM=180gxmol^-1), dissolved in 72g of H2O? The vapor pressure of pure water at 25Celsius is 23.8 torr.
vp pure solvent - lowering due to glucose = vp pure solvent - (mole fraction glucose x vp pure solvent) =
P<sup>o</sup>- X<sub>glucose</sub>P<sup>o</sup>
mols glucose = 45/180 = 0.25
mols water = 72/18 = 4
total mols = 4.25
mols fraction glucose = X = 0.25/4.25 = 0.588
P<sup>o</sup> = 23.8
Note I omitted a zero.
That is 0.25/4.25 = 0.0588
thanks, you're really good at putting this in english
vp pure solvent - lowering due to glucose = vp pure solvent - (mole fraction glucose x vp pure solvent) =
P<sup>o</sup>- X<sub>glucose</sub>P<sup>o</sup>
mols glucose = 45/180 = 0.25
mols water = 72/18 = 4
total mols = 4.25
mols fraction glucose = X = 0.25/4.25 = 0.588
P<sup>o</sup> = 23.8
Note I omitted a zero.
That is 0.25/4.25 = 0.0588
thanks, you're really good at putting this in english