Question

A gas mixture of nitrogen and oxygen having a total pressure of 2.50 atm is above 2.0 L of water at 25 °C. The water has 51.3 mg of nitrogen dissolved in it. What is the molar composition of nitrogen and oxygen in the gas mixture? The Henry’s constants for N2 and O2 in water at 25 °C are 6.1×10–4 M/atm and 1.3×10–3 M/atm, respectively.

Answer 1

Henry's law is

C = K P

where,

C = Concentration of gas in liquid

K = Henry's Law Constant

P = Partial pressure of gas above liquid

mass of dissolved Nitrogen = 0.0513g

No of mole of Nitrogen gas dissolved = 0.0513g/28.0134g/mol=0.00183127mole

Concentration of Nitrogen gas in water =(0.00183127mole/2L)*1L= 0.0009156M

Applying the Henry's law

0.0009156M=6.1*10^-4M/atm *P

P = 0.0009156 M/6.1*10^-4M/atm

= 1.50 atm

So, Partial pressure of N2 in gas mixture = 1.5atm

Total pressure = 2.50atm

Partial pressure of O2 = 2.50atm - 1.50atm = 1atm

Pressure fraction of N2=1.50atm/2.50atm=0.6

Pressure fraction of O2 = 1atm/2.50atm = 0.4

Mole fraction = pressure fraction

Therefore,

In gas mixture

Mole fraction of N2 = 0.60

Mole fraction of O2 = 0.40