What is the vapor pressure at 23 degrees C of a solution of 1.20 g of naphthalene, C10H8, in 25.6 g of benzene, C6H6? The vapor pressure of pure benzene at 23 degrees C is 86.0 mmHg; the vapor pressure of naphthalene can be neglected. Calcluate the vapor pressure lowering of the solution.

To calculate the vapor pressure lowering of a solution, we can use Raoult's law, which states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution.

To start with, we need to determine the number of moles of naphthalene and benzene in the solution.

1. Calculate the number of moles of naphthalene (C10H8):
- We have 1.20 g of naphthalene.
- The molar mass of naphthalene is 128.17 g/mol.
- Divide the mass by the molar mass to get the number of moles:
Moles of naphthalene = 1.20 g / 128.17 g/mol

2. Calculate the number of moles of benzene (C6H6):
- We have 25.6 g of benzene.
- The molar mass of benzene is 78.11 g/mol.
- Divide the mass by the molar mass to get the number of moles:
Moles of benzene = 25.6 g / 78.11 g/mol

Next, we can calculate the vapor pressure lowering of the solution using Raoult's law.

3. Calculate the mole fraction of benzene (Xbenzene):
- The mole fraction is the moles of benzene divided by the total moles of solute and solvent:
Xbenzene = Moles of benzene / (Moles of benzene + Moles of naphthalene)

4. Calculate the vapor pressure lowering (ΔP):
- According to Raoult's law, ΔP = Xbenzene * Pbenzene, where Pbenzene is the vapor pressure of pure benzene at the given temperature.
ΔP = Xbenzene * Pbenzene

Substitute the given values to calculate the vapor pressure lowering:

- The vapor pressure of pure benzene (Pbenzene) is given as 86.0 mmHg.
- We've already calculated the mole fraction of benzene (Xbenzene) using the moles calculated in step 2.

Plug in the values and calculate the vapor pressure lowering.

delta Psolvent = XsolutePosolvent