Answer: The solubility of nitrogen gas at 5.00 atm is [tex]117.6mg/100g[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{A}=K_H\times p_{A}[/tex]
Or,
[tex]\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}[/tex]
where,
[tex]C_1\text{ and }p_1[/tex] are the initial concentration and partial pressure of nitrogen gas
[tex]C_2\text{ and }p_2[/tex] are the final concentration and partial pressure of nitrogen gas
We are given:
[tex]C_1=56.0mg/100g\\p_1=2.38atm\\C_2=?\\p_2=5.00atm[/tex]
Putting values in above equation, we get:
[tex]\frac{56.0mg/100g}{C_2}=\frac{2.38atm}{5.00atm}\\\\C_2=\frac{56.0mg/100g\times 5.00atm}{2.38atm}=117.6mg/100g[/tex]
Hence, the solubility of nitrogen gas at 5.00 atm is [tex]117.6mg/100g[/tex]
