Question

A solution is prepared by dissolving 40.0g of sucrose, C₁₂H₂₂O₁₁, in 250.0 g of water at 25 degrees C. What is the vapor pressure of the solution if the vapor pressure of water at 25 degrees C is 23.76 mm Hg?

Answer 1

To find the vapor pressure of the solution, we can use Raoult's Law. According to Raoult's Law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

Step-by-step explanation:

To find the vapor pressure of the solution, we need to use Raoult's Law. According to Raoult's Law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

In this case, the solvent is water and the solute is sucrose. We can first calculate the mole fraction of water:

Mole fraction of water = moles of water / total moles of solute and solvent

To find the moles of water, we can use the formula:

Moles = Mass / Molar mass

The molar mass of water is 18.01528 g/mol, so:

Moles of water = 250.0 g / 18.01528 g/mol

The moles of sucrose can be calculated the same way:

Moles of sucrose = 40.0 g / 342.29648 g/mol

The total moles of water and sucrose are the sum of the moles of water and moles of sucrose.

Finally, we can calculate the mole fraction of water and use it to find the vapor pressure of the solution:

Vapor pressure of solution = (mole fraction of water) * (vapor pressure of pure water)