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

Answer: The molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of nitrogen gas = 51.3 mg = 0.0513 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of nitrogen gas = 28 g/mol

Volume of solution = 2 L

Putting values in above equation, we get:

$\text{Molarity of nitrogen gas}=\frac{0.0513g}{28g/mol\times 2L}\\\\\text{Molarity of nitrogen gas}=9.16\times 10^{-4}mol/L$

To calculate the partial pressure, we use the equation given by Henry's law, which is:

$C_{N_2}=K_H\times p_{N_2}$

where,

$K_H$ = Henry's constant = $6.1\times 10^{-4}mol/L.atm$

$C_{N_2}$ = molar solubility of nitrogen gas = $9.16\times 10^{-4}mol/L$

Putting values in above equation, we get:

$9.16\times 10^{-4}mol/L=6.1\times 10^{-4}mol/L.atm\times p_{N_2}\\\\p_{N_2}=\frac{9.16\times 10^{-4}mol/L}{6.1\times 10^{-4}mol/L.atm}=1.50atm$

We are given:

Total pressure of the mixture = 2.50 atm

Partial pressure of oxygen gas = 2.50 - 1.50 = 1.00 atm

To calculate the mole fraction of a substance at 25°C, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$       ......(1)

where,

$p_A$ = partial pressure

$p_T$ = total pressure

$\chi_A$ = mole fraction

• For nitrogen gas:

We are given:

$p_{N_2}=1.50atm\\p_T=2.50atm$

Putting values in equation 1, we get:

$1.50atm=2.50\times \chi_{N_2}\\\\\chi_{N_2}=\frac{1.50}{2.50}=0.6$

• For oxygen gas:

We are given:

$p_{O_2}=1.00atm\\p_T=2.50atm$

Putting values in equation 1, we get:

$1.00atm=2.50\times \chi_{O_2}\\\\\chi_{O_2}=\frac{1.00}{2.50}=0.4$

Hence, the molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4