Question

Calculate the mass of nitrogen dissolved at room temperature in80.0 L home aquarium. Assume a total pressure of 1.0 atm and a molefraction for nitrogen of 0.78.

Answer 1

I thought that was the right process, but this is actually completely wrong. What the person above found was the pressure of the gas above the liquid or aquarium instead of the gas actually DISSOLVED in the aquarium.

To find this number you must use Henry's law which is:

Sgas=(Kh)(Pgas)

where:

*Kh is the Henry's constant in units of M/atm, it can be found in a chemistry book or online.

*Pgas is the PARTIAL pressure of the nitrogen gas in units of atm

 

in this case Kh for nitrogen is 6.1x10^-4 M/atm

and specifically for this case, the total pressure is equal to the partial pressure of nitrogen because

Partial pressure of nitrogen= mole fraction x total pressure

= .78 x 1

= .78 atm

Plug in your values into Henry's law ( Kh=6.1*10^-4 M/atm, and Pgas=.78 atm) and you get a solubility (Sgas) of

4.76x 10 ^-4 M

using your molarity equation multiply this number (4.76x 10 ^-4 M) by the liters which is 80 L and

you get .038 moles.

 

using basic stoichiometry multiply this number by the molar mass of nitrogen (28g/mol) and

you get a total of 1.1 g of Nitrogen gas dissolved in the aquarium.

Answer 2

P_total = 1 atm
mole fraction of Nitrogen = .78
Partial pressure of Nitrogen = P_N2 = Total pressure xmole fraction of nitrogen = 1 atm x .78 = .78 atm

V = 80.0 L
Temperature = 25 C (room temperature) = 298 K

PV = nRT
n = PV/RT
n = (.78 atm)(80.0 L) / (.0821)(298)
n = 2.55 mol

Molar mass N2 = 28 g/mol
Mass of N2 dissolved in 80 L aquarium = moles N2 x Molar massN2
Mass dissolved = 2.55 x 28 = 71.4 g