To find the vapor pressure of a 66.0% aqueous solution of sucrose at 75.0 °C, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is proportional to the mole fraction of the solvent.
First, we need to calculate the mole fraction of water in the solution. Since the solution is 66.0% aqueous, the other 34.0% must be sucrose.
To calculate the mole fraction of water, we can use the following equation:
mole fraction of water = moles of water / total moles of solute
Since we know the mass percent of sucrose, we can assume we have 100 g of solution. The mass of water in the solution can be calculated as follows:
mass of water = 66.0% * 100 g = 66.0 g
To find the moles of water, we can divide the mass by the molar mass of water (18.015 g/mol):
moles of water = 66.0 g / 18.015 g/mol = 3.662 mol
The mass of sucrose can be calculated as:
mass of sucrose = 34.0% * 100 g = 34.0 g
To find the moles of sucrose, we can divide the mass by the molar mass of sucrose (342.30 g/mol):
moles of sucrose = 34.0 g / 342.30 g/mol = 0.099 mol
The total moles of solute in the solution is the sum of the moles of water and moles of sucrose:
total moles of solute = moles of water + moles of sucrose = 3.662 mol + 0.099 mol = 3.761 mol
Now, we can calculate the mole fraction of water:
mole fraction of water = moles of water / total moles of solute = 3.662 mol / 3.761 mol = 0.974
Since the solution obeys Raoult's law, the vapor pressure of the solution is proportional to the mole fraction of water. Therefore, the vapor pressure of the solution is 0.974 times the vapor pressure of pure water at 75.0 °C.
The vapor pressure of pure water at 75.0 °C is given as 380 torr. Therefore, the vapor pressure of the 66.0% aqueous solution of sucrose is:
vapor pressure = 0.974 * 380 torr = 370.47 torr
To convert torr to atm, we divide by 760:
vapor pressure = 370.47 torr / 760 torr/atm = 0.487 atm
Therefore, the vapor pressure in atm of the 66.0% aqueous solution of sucrose at 75.0 °C is approximately 0.487 atm.