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.
oooh. thank you. :-)
To determine the vapor pressure of the pure solvent, we need to use Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.
First, we need to find the mole fraction of the solvent in the mixture. The mole fraction is calculated by dividing the number of moles of the solvent by the total number of moles in the solution:
Mole fraction of solvent = Moles of solvent / Total moles
In this case, the moles of solvent are given as 0.8 mol, and the total moles in the solution are the sum of the moles of solvent and solute, which is 0.8 mol + 0.4 mol = 1.2 mol.
Mole fraction of solvent = 0.8 mol / 1.2 mol = 0.6667
Since the solute is nonvolatile, it does not contribute to the vapor pressure. Therefore, the mole fraction of the solvent in the pure solvent is also 0.6667.
According to Raoult's Law, the vapor pressure of the pure solvent is equal to the mole fraction of the solvent multiplied by the vapor pressure of the solvent in the mixture.
Vapor pressure of pure solvent = Mole fraction of solvent * Vapor pressure of solvent in mixture
Vapor pressure of pure solvent = 0.6667 * 60 mmHg = 40 mmHg
Therefore, the vapor pressure that should be reported at 300 K for the pure solvent is 40 mmHg.
The correct answer is not B) 60 mmHg, but rather A) 40 mmHg.