the vapor pressure of carbon sulfide is 355.6 torr at 25 degrees Celsius. what is the vapor pressure of a solution prepared by dissolving 12.52 g naphthalene (C10H8, 128.2 g/mol) in 125.0 mL CS2 liquid (76.14 g/mol, 1.261 g/mL)? Assume the solution obeys Raoult's law, and treat naphthalene as a nonvolatile solute. what is the torr?
To find the vapor pressure of the solution, we need to use Raoult's Law, which states that the vapor pressure of a solvent above a solution is equal to the mole fraction of the solvent multiplied by its pure vapor pressure.
First, let's calculate the mole fraction of the solvent (CS2) in the solution:
1. Calculate the moles of CS2:
mass of CS2 = volume of CS2 x density of CS2
mass of CS2 = 125.0 mL x 1.261 g/mL
mass of CS2 = 157.63 g
moles of CS2 = mass of CS2 / molar mass of CS2
moles of CS2 = 157.63 g / 76.14 g/mol
moles of CS2 = 2.068 mol
2. Calculate the mole fraction of CS2:
mole fraction of CS2 = moles of CS2 / (moles of CS2 + moles of naphthalene)
mole fraction of CS2 = 2.068 mol / (2.068 mol + 0 mol)
mole fraction of CS2 = 1.0
Since naphthalene is treated as a nonvolatile solute, its contribution to the vapor pressure is negligible. Therefore, the mole fraction of naphthalene is zero in this case.
Now, we can find the vapor pressure of the solution:
vapor pressure of the solution = mole fraction of CS2 x vapor pressure of pure CS2
Since the mole fraction of CS2 is 1.0 (as calculated above) and the vapor pressure of pure CS2 is given as 355.6 torr, we have:
vapor pressure of the solution = 1.0 x 355.6 torr
vapor pressure of the solution = 355.6 torr
So, the vapor pressure of the solution prepared by dissolving naphthalene in CS2 is 355.6 torr.
Convert g each to mols and convert to XCS2
Then psolution = XCS2*Po CS2
Post your work if you get stuck and I can help you through it.