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?
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.