Gen Chem, Factors Affecting Solubility

[coffeecat91]

Homework Statement

Calculate the mass of nitrogen dissolved at room temperature in an 80.0L home aquarium. Assume a total pressure of 1.0 atm and a mole fraction for nitrogen of .78.

Homework Equations

Mole fraction= moles solute/moles solution

S_gas=K_H x P_gas

K_H = 6.1 x 10^-4 M/atm

The Attempt at a Solution

S_gas = (6.1 x 10^-4)(1 atm)
S_gas = 6.1 x 10^-4
6.1 x 10^-4/ 1 L x 80.0 L = 4.88 x 10^-2 moles
(4.88 x 10^-2 moles) x 28.02 g N_s/ 1 mol = 1.37 g N₂

the correct answer is 1.1 g N₂
Where did I go wrong and where does the mole fraction bit come into play? Thanks!

 

[Borek]

Where did I go wrong and where does the mole fraction bit come into play?

Partial pressure.

 

[Blueshift5]

Your calculation for the mass of nitrogen dissolved is correct. The mole fraction comes into play when calculating the solubility of a gas in a liquid. In this case, the mole fraction of nitrogen in the solution is 0.78, meaning that 78% of the moles in the solution are nitrogen. This is important because the solubility of a gas is directly proportional to its mole fraction in the solution. So, if we use the mole fraction of 0.78 instead of 1 (for the total moles in the solution), we get a slightly different answer:

Sgas = (6.1 x 10-4)(1 atm) = 6.1 x 10-4
6.1 x 10-4/ 1 L x 80.0 L = 4.88 x 10-2 moles
(4.88 x 10-2 moles) x 0.78 = 3.81 x 10-2 moles N2
(3.81 x 10-2 moles) x 28.02 g Ns/ 1 mol = 1.07 g N2

This is closer to the correct answer of 1.1 g N2. So, the mole fraction is important in accurately calculating the solubility of a gas in a liquid. Additionally, it is important to note that the solubility of a gas is also affected by temperature and pressure, which can be taken into account using the Henry's Law equation (Sgas = KH x Pgas).