<u>Answer:</u> The molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4
<u>Explanation:</u>
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)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
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](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20nitrogen%20gas%7D%3D%5Cfrac%7B0.0513g%7D%7B28g%2Fmol%5Ctimes%202L%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20nitrogen%20gas%7D%3D9.16%5Ctimes%2010%5E%7B-4%7Dmol%2FL)
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}](https://tex.z-dn.net/?f=C_%7BN_2%7D%3DK_H%5Ctimes%20p_%7BN_2%7D)
where,
= Henry's constant = ![6.1\times 10^{-4}mol/L.atm](https://tex.z-dn.net/?f=6.1%5Ctimes%2010%5E%7B-4%7Dmol%2FL.atm)
= molar solubility of nitrogen gas = ![9.16\times 10^{-4}mol/L](https://tex.z-dn.net/?f=9.16%5Ctimes%2010%5E%7B-4%7Dmol%2FL)
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](https://tex.z-dn.net/?f=9.16%5Ctimes%2010%5E%7B-4%7Dmol%2FL%3D6.1%5Ctimes%2010%5E%7B-4%7Dmol%2FL.atm%5Ctimes%20p_%7BN_2%7D%5C%5C%5C%5Cp_%7BN_2%7D%3D%5Cfrac%7B9.16%5Ctimes%2010%5E%7B-4%7Dmol%2FL%7D%7B6.1%5Ctimes%2010%5E%7B-4%7Dmol%2FL.atm%7D%3D1.50atm)
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:
......(1)
where,
= partial pressure
= total pressure
= mole fraction
We are given:
![p_{N_2}=1.50atm\\p_T=2.50atm](https://tex.z-dn.net/?f=p_%7BN_2%7D%3D1.50atm%5C%5Cp_T%3D2.50atm)
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](https://tex.z-dn.net/?f=1.50atm%3D2.50%5Ctimes%20%5Cchi_%7BN_2%7D%5C%5C%5C%5C%5Cchi_%7BN_2%7D%3D%5Cfrac%7B1.50%7D%7B2.50%7D%3D0.6)
We are given:
![p_{O_2}=1.00atm\\p_T=2.50atm](https://tex.z-dn.net/?f=p_%7BO_2%7D%3D1.00atm%5C%5Cp_T%3D2.50atm)
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](https://tex.z-dn.net/?f=1.00atm%3D2.50%5Ctimes%20%5Cchi_%7BO_2%7D%5C%5C%5C%5C%5Cchi_%7BO_2%7D%3D%5Cfrac%7B1.00%7D%7B2.50%7D%3D0.4)
Hence, the molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4