An aqueous solution has a vapor pressure of 81.1 at 50 . The vapor pressure of pure water at this temperature is 92.6. What is the concentration of in mass percent?
What is the concn of WHAT? in mass percent?
Psoln = Xsolvent*Psolvent
81.1 = Xsolvent*(92.6)
Solve for X which is the mole fraction of the solvent,
1-mole fraction solvent = mole fraction solute.
Convert mole fraction solute to mass percent.
To determine the concentration of a solute in mass percent, we need to compare the vapor pressure of the solution with the vapor pressure of the pure solvent at the same temperature.
The vapor pressure is a colligative property, meaning it depends on the number of solute particles rather than their identity. According to Raoult's Law, the vapor pressure of a solution is directly proportional to the mole fraction of the solute in the solution.
Let's assume the solute in this case is glucose (C6H12O6). We can set up the equation using Raoult's Law:
Vapor pressure of solution = (mole fraction of glucose) × (vapor pressure of pure water)
Since we are looking for the concentration in mass percent, we need to convert the mass of the solute (glucose) into moles.
Now, let's plug in the given values:
Vapor pressure of solution = 81.1 mmHg
Vapor pressure of pure water = 92.6 mmHg
Rearranging the equation, we get:
(mole fraction of glucose) = (vapor pressure of solution) / (vapor pressure of pure water)
Substituting the given values:
(mole fraction of glucose) = 81.1 mmHg / 92.6 mmHg
(mole fraction of glucose) = 0.8768
To convert the mole fraction into mass percent, we need to use the molar mass of glucose, which is 180.16 g/mol.
Mass percent of glucose = (moles of glucose × molar mass of glucose) / (mass of solution) × 100%
Since we don't know the mass of the solution, we can assume it to be 100 grams for simplicity.
Mass percent of glucose = (0.8768 moles × 180.16 g/mol) / 100 g × 100%
Mass percent of glucose = 157.83%
Therefore, the concentration of glucose in the solution is approximately 157.83% by mass.