Calculate the vapor pressure of water above the solution.?
The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation of the solute.
The density of a 1.50 M solution of sodium nitrate (NaNO3) at 298 K is 1.021 g/mL.
I have no idea how to do this.
- ChemTeamLv 71 month agoFavourite answer
We are going to use Raoult's Law but we have to do some preparation first. This is because we need to know the mole fraction of the solvent. Some people solve these problems by getting the mole fraction of the solute.
We start by assuming 1.00 L of solution is present. That tells us two things:
1) there are 1.50 moles of NaNO3 present
2) the total weight of the solution is 1021 g.
I need to know how much of the solution is water and how much is NaNO3.
(1.50 mol) (84.994 g/mol) = 127.491 g <--- mass of NaNO3 in the 1.00 L of solution
1021 g - 127.491 g = 893.509 g <--- mass of water in the 1.00 L of solution
We need the moles of each:
1.50 mol <--- a twist is coming up regarding this number.
893.509 g / 18.015 g/mol = 49.598 mol (of water in the 1.00 L)
Here's the twist:
Colligative properties are based on the number of particles in solution. Since NaNO3 dissociates in solution, that means we have 3.00 mol of little things (Na^+ ions and NO3^- ions) floating around in solution.
This adjustment is called the van 't Hoff factor and I will leave you to study it more.
Now, the mole fraction of water:
49.598 / (49.598 + 3.00) = 0.9429636 <--- no units and I'll keep some extra digits)
Finally, we use Raoult's Law:
P_solution = (mole fraction solvent) (P_pure solvent)
P_solution = (0.9429636) (0.0313 atm)
P_solution = 0.0295 atm
If you calculated the mole fraction of the solute, then Raoult's Law would give you the amount by which the vapor pressure of the solution decreased. Then, you'd subtract that value from 0.0313 atm for the final answer.
"I have no idea how to do this."
I hope I was able to change that! Best wishes in your studies.