contestada

What is the solubility of ethylene (in units of grams per liter) in water at 25 °C, when the C2H4 gas over the solution has a partial pressure of 0.684 atm? kH for C2H4 at 25 °C is 4.78×10-3 mol/L·atm.

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]