Question

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.

Answer 1

Answer: The solubility of ethylene gas in water is $9.16\times 10^{-2}g/L$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{C_2H_4}=K_H\times p_{C_2H_4}$

where,

$K_H$ = Henry's constant = $4.78\times 10^{-3}mol/L.atm$

$C_{C_2H_4}$ = molar solubility of ethylene gas = ?

$p_{C_2H_4}$ = partial pressure of ethylene gas = 0.684 atm

Putting values in above equation, we get:

$C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L$

Converting this into grams per liter, by multiplying with the molar mass of ethylene:

Molar mass of ethylene gas = 28 g/mol

So, $C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L$

Hence, the solubility of ethylene gas in water is $9.16\times 10^{-2}g/L$