A solution is prepared by dissolving 248 g sucrose (C12H22O11(s); MM = 342 g/mol) in 452 g H2O(l) (MM = 18.0 g/mol). What is the Vapor Pressure of this solution if the pure vapor pressure of water at 30 0C is 31.8 mmHg?

1 answer

To find the vapor pressure of the solution, we need to use Raoult's Law, which states that the vapor pressure of a solution is proportional to the mole fraction of the solvent present.

First, we need to calculate the moles of sucrose and water in the solution.

Moles of sucrose = mass of sucrose / molar mass of sucrose
Moles of sucrose = 248 g / 342 g/mol
Moles of sucrose = 0.725 mol

Moles of water = mass of water / molar mass of water
Moles of water = 452 g / 18.0 g/mol
Moles of water = 25.1 mol

Next, we need to calculate the mole fraction of water in the solution.

Mole fraction of water = moles of water / total moles
Mole fraction of water = 25.1 mol / (0.725 mol + 25.1 mol)
Mole fraction of water = 0.971

Now we can use Raoult's Law to calculate the vapor pressure of the solution.

Vapor pressure of solution = mole fraction of solvent * pure vapor pressure of solvent
Vapor pressure of solution = 0.971 * 31.8 mmHg
Vapor pressure of solution = 30.88 mmHg

Therefore, the vapor pressure of the solution is 30.88 mmHg.