an ideal mixture contains 0.4 mol of nonvolatile solute and 0.8 mol of volatile solvent. the resulting vapor pressure was 60 mmHg at 300K. What vapor pressure should be reported at 300 K for pure solvent?
A) 48 mmHg
B) 60 mmHg
C) 75 mmHg
D) 90 mmHg
**Would it be B) 60 mmHg since the solute is nonvolatile?
I don't think so. When Raoult's Law tells you that the vapor pressure of a pure solvent is lowered when a non-volatile solvent is added.
mols solute = 0.4
mols solvent = 0.8
total mols 1.2
Xsolvent = 0.8/1.2 = ?
Then pmixture = Xsolvent*Po solvent. So it must be either 75 or 90. Work it out.posted by DrBob222
oooh. thank you. :-)posted by Anonymous