A solution is prepared by adding 1g of sugar mannitol (MW 182.2g/g-mole) in 100g of water at 20◦C. (a) Assuming that mannitol and water form an ideal solution, what is the equilibrium vapor pressure of solution? The vapor pressure of pure water is 17.54mmHg at 20◦C
mols manitol = grams/molar mass = ?
mols H2O = grams/molar mass = ?
Find total mols by adding.
Then find mol fraction H2O = mols H2O/total mols.
Then Psoln = XH2O*Popure H2O
To determine the equilibrium vapor pressure of the solution, we need to apply Raoult's Law. Raoult's Law states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent.
First, let's calculate the mole fraction of water.
Mole fraction (Xwater) = moles of water / total moles in solution
Moles of water = mass of water / molar mass of water
= 100 g / 18.015 g/mol ≈ 5.548 mol
Moles of mannitol = mass of mannitol / molar mass of mannitol
= 1 g / 182.2 g/mol ≈ 0.00549 mol
Total moles in solution = moles of water + moles of mannitol
= 5.548 mol + 0.00549 mol ≈ 5.553 mol
Mole fraction of water (Xwater) = 5.548 mol / 5.553 mol ≈ 0.99892
Since mannitol is a non-volatile solute, its contribution to the vapor pressure is negligible. Therefore, we can assume that the mole fraction of water is the same as the mole fraction of the solvent in the vapor phase, which is equal to 0.99892.
Now, we can apply Raoult's Law to calculate the equilibrium vapor pressure of the solution.
Psolution = Xwater * Pwater
Pwater = 17.54 mmHg (given)
Psolution = 0.99892 * 17.54 mmHg ≈ 17.531 mmHg
Therefore, the equilibrium vapor pressure of the solution is approximately 17.531 mmHg.