Compute the vapor pressure of water above a solution formed by dissolving 25.00 grams of magnesium sulfate, MgSO4 (MM=120.3) in 95.00 grams of water at 338K. The vapor pressure of pure water at 338K is 188.0 torr.
P(solvent) = X(solvent)* P(normal v.p. solvent).
To compute the vapor pressure of water above a solution, formed by dissolving a solute in water, we can use Raoult's law. Raoult's law states that the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
Here are the steps to solve the problem:
Step 1: Calculate the number of moles of each component in the solution.
The number of moles of magnesium sulfate (MgSO4) can be calculated using the given mass and molar mass:
25.00 grams of MgSO4 * (1 mole / 120.3 grams) = 0.2079 moles of MgSO4
The number of moles of water can be calculated using the given mass and molar mass:
95.00 grams of water * (1 mole / 18.015 grams) = 5.2755 moles of water
Step 2: Calculate the mole fraction of water in the solution.
The mole fraction (X) of water is given by the moles of water divided by the total moles in the solution:
X(water) = moles of water / (moles of water + moles of MgSO4)
X(water) = 5.2755 moles / (5.2755 moles + 0.2079 moles) = 0.9623
Step 3: Calculate the vapor pressure of water above the solution.
The vapor pressure of water above the solution can be calculated using Raoult's law:
P(water) = X(water) * P(water pure)
P(water) = 0.9623 * 188.0 torr = 181.1336 torr
Therefore, the vapor pressure of water above the solution formed by dissolving 25.00 grams of magnesium sulfate (MgSO4) in 95.00 grams of water at 338K is approximately 181.1336 torr.