By how much will the vapor pressure of water decrease if a solution at 40.0°C is produced by dissolving 16.0212 g of K3PO4 (formula mass = 212.30 g/mol) with 40.8480 g of water. The vapor pressure of pure water at this temperature is 55.30 mmHg. If the solute is ionic, assume that it is completely ionic!
we use Raoults law but i don't know how to use the g given
mols K3PO4 = grams/molar mass
mols H2O = grams/molar mass
Total mols = ?
XK3PO4 = ?
XH2O = ?
NOTE. Remember i, the van't Hoff factor for H3PO4 is 3and take that into account in the above, or at least before calculate ing vapor pressure.
To solve this problem, we can use Raoult's Law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
First, we need to calculate the mole fraction of water in the solution. To do this, we need to determine the number of moles of water and the number of moles of K3PO4.
Number of moles of water = mass of water / molar mass of water
Number of moles of water = 40.8480 g / (18.015 g/mol) = 2.2684 mol
Number of moles of K3PO4 = mass of K3PO4 / molar mass of K3PO4
Number of moles of K3PO4 = 16.0212 g / (212.30 g/mol) = 0.0755 mol
Next, we calculate the mole fraction of water:
Mole fraction of water = moles of water / (moles of water + moles of K3PO4)
Mole fraction of water = 2.2684 mol / (2.2684 mol + 0.0755 mol) = 0.9678
Now, we can use Raoult's Law to calculate the vapor pressure of the solution:
Vapor pressure of water in solution = Vapor pressure of pure water * Mole fraction of water
Vapor pressure of water in solution = 55.30 mmHg * 0.9678 = 53.53 mmHg
Therefore, the vapor pressure of water in the solution at 40.0°C will decrease by 53.53 mmHg compared to its vapor pressure as a pure substance.