Answer: The solubility of nitrogen gas in the sample is [tex]5.6\times 10^{-4}mol/L[/tex]
Explanation:
We are given:
Volume percent of nitrogen gas = 80 %
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{N_2}=K_H\times p_{N_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]7.0\times 10^{-4}mol/L.atm[/tex]
[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas = 1.0 atm
Putting values in above equation, we get:
[tex]C_{N_2}=7.0\times 10^{-4}mol/L.atm\times 1.0\\\\C_{N_2}=7.0\times 10^{-4}mol/L[/tex]
Solubility of nitrogen gas in the sample = [tex]\frac{80}{100}\times 7.0\times 10^{-4}mol/=5.6\times 10^{-4}mol/L[/tex]
Hence, the solubility of nitrogen gas in the sample is [tex]5.6\times 10^{-4}mol/L[/tex]
