What is the vapor pressure of a solution prepares by dissolving 4.25g of K2O in enough water to preapare 1.25L of Solution at 25 degrees (C)? Assume that the density of the solution is 1.00 g/mL and that the vapor pressure of pure water is 23.8 torr at 25 degrees (C).
I've gotten the answer because the teacher went over it in class. But it makes no sense to me
Answer: 23.3 torr
To calculate the vapor pressure of a solution, you need to use Raoult's law. Raoult's law states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution.
1. Convert the mass of K2O to moles:
- The molar mass of K2O is 94.20 g/mol.
- Divide the mass (4.25g) by the molar mass to get the number of moles: 4.25g / 94.20 g/mol = 0.045 moles.
2. Calculate the mole fraction of the solvent (water):
- To do this, you need to know the moles of water present in the solution.
- Since the density of the solution is given as 1.00 g/mL, the mass of the solution is equal to its volume (1.25L or 1250g).
- Subtract the mass of K2O from the total mass of the solution to find the mass of water: 1250g - 4.25g = 1245.75g.
- Convert the mass of water to moles. The molar mass of water (H2O) is 18.02 g/mol.
- Divide the mass (1245.75g) by the molar mass to get the number of moles: 1245.75g / 18.02 g/mol = 69.09 moles.
- The mole fraction of water is the moles of water divided by the total moles: 69.09 moles / (69.09 moles + 0.045 moles) = 0.9994.
3. Calculate the vapor pressure using Raoult's law:
- Since the vapor pressure of pure water is given as 23.8 torr, multiply it by the mole fraction of water: 23.8 torr * 0.9994 = 23.77 torr.
Therefore, the vapor pressure of the solution prepared by dissolving 4.25g of K2O in enough water to prepare 1.25L of solution at 25 degrees (C) is approximately 23.8 torr.