Vapor Pressure Help Needed!

Question

The density of a 1.50 M solution of sodium nitrate (NaNO3) at 298 K is 1.021 g/mL.

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.

Assistance in setting it up would be helpful!

Answer 1

Refer to Raoult's Law

When a solute such as NaNO₃ is added to water, it will LOWER the vapor pressure of the pure water. The amount of lowering is dependent on the concentration of the solute and is directly related to the mole fraction of the solute. So, we will find the moles of NaNO₃, the moles of H₂O, total moles, and mole fraction of H₂O.

We have a density of 1.021 g /ml. Let's assume we have 1000 ml. That means we have...

1.021 g/ml x 1000 ml = 1021 g

Since the concentration of NaNO₃ is 1.5 M, that means 1.5 moles per liter. Since we have 1000 ml,( 1 L ) so we have 1.5 moles NaNO₃. How many grams is that? Molar mass NaNO₃ = 84.99 g/mol

1.5 moles NaNO₃ x 84.99 g/mol = 127.5 g NaNO₃

Mass of water = 1021 g - 127.5 g = 893.5 g H₂O

Moles NaNO₃ = 1.5 moles NaNO₃

Moles H2O = 893.5 g H₂O x 1 mol / 18 g = 49.6 moles H₂O

Total moles = 1.5 + 49.6 = 51.1 moles total

Mole fraction H₂O = 49.6 mol / 51.1 mol = 0.971 and according to Raoult's law...

vapor pressure of solution = mole fraction of solvent x vapor pressure of pure solvent ...

Vapor pressure of solution = 0.971 x 0.0313 atm = 0.0304 atm (3 sig. figs.)