Question

If a gaseous sample contains 80.% N2 by volume, what is the solubility of N2 in water at 25°C and 1.0 atm (kH in H2O at 25°C = 7.0 10-4 mol/Latm)?

Answer 1

Answer: The solubility of nitrogen gas in the sample is $5.6\times 10^{-4}mol/L$

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:

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

where,

$K_H$ = Henry's constant = $7.0\times 10^{-4}mol/L.atm$

$p_{N_2}$ = partial pressure of nitrogen gas = 1.0 atm

Putting values in above equation, we get:

$C_{N_2}=7.0\times 10^{-4}mol/L.atm\times 1.0\\\\C_{N_2}=7.0\times 10^{-4}mol/L$

Solubility of nitrogen gas in the sample = $\frac{80}{100}\times 7.0\times 10^{-4}mol/=5.6\times 10^{-4}mol/L$

Hence, the solubility of nitrogen gas in the sample is $5.6\times 10^{-4}mol/L$