Respuesta :
Answer: The solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{C_2H_4}=K_H\times p_{C_2H_4}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]4.78\times 10^{-3}mol/L.atm[/tex]
[tex]C_{C_2H_4}[/tex] = molar solubility of ethylene gas = ?
[tex]p_{C_2H_4}[/tex] = partial pressure of ethylene gas = 0.684 atm
Putting values in above equation, we get:
[tex]C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L[/tex]
Converting this into grams per liter, by multiplying with the molar mass of ethylene:
Molar mass of ethylene gas = 28 g/mol
So, [tex]C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L[/tex]
Hence, the solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]