2.8 g of the nonvolatile solute sucrose (C12H22O11) is added to 517 g of water at 25oC. What will be the partial pressure of the water vapor over this solution?
Isn't partial pressure =
P = Po*Xo ?
So convert g of solvent to mole fraction and multiply by the normal vapor pressure of the solvent. Post your work if you get stuck.
So 2.8 converted to mols and then divided by 517? And multiplied by 25oC?
No.
mole fraction of solvent =
moles solvent/(moles solvent + moles solute)
To find the partial pressure of water vapor over the solution, we need to use Raoult's law. According to Raoult's law, the partial pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.
First, we need to determine the mole fraction of water in the solution.
To find the mole fraction, we need to calculate the moles of sucrose and water in the solution.
The molar mass of sucrose (C12H22O11) is:
12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.34 g/mol
Given that the mass of sucrose is 2.8 g, we can calculate the moles of sucrose:
moles of sucrose = mass of sucrose / molar mass of sucrose
moles of sucrose = 2.8 g / 342.34 g/mol
Next, we need to calculate the moles of water:
moles of water = mass of water / molar mass of water
moles of water = 517 g / 18.02 g/mol
Now, we can calculate the mole fraction of water:
mole fraction of water = moles of water / (moles of water + moles of sucrose)
After finding the mole fraction of water, we need to find the vapor pressure of pure water at 25°C. For this, we can consult a reference source or use the Clausius-Clapeyron equation.
Finally, we can calculate the partial pressure of water using Raoult's law:
partial pressure of water vapor = mole fraction of water * vapor pressure of pure water
Once you find the mole fraction of water and the vapor pressure of pure water, you can multiply them to find the partial pressure of the water vapor over the solution.