Question

At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubility of N2 gas in water, at the same temperature, if the partial pressure of N2 gas over the solution is increased from 4.07 atm to 10.0 atm .

Answer 1

Answer: Thus the solubility of $N_2$ gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of $N_{2}$ in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  $C=k_{H}P$.

$k_{H}$= Henry’s law constant = ?

The partial pressure (P) of $N_{2}$ in water = 4.07 atm

\$C= k_{H}\times P\\95.7mg=k_{H}\times 4.07$

$k_{H}=23.5$

At pressure of 10.0 atm

$C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg$

Thus the solubility of $N_2$ gas in water, at the same temperature, is 235mg/100g