Question

What is the solubility of argon (in units of grams per liter) in water at 25 °C, when the Ar gas over the solution has a partial pressure of 0.430 atm? kH for Ar at 25 °C is 1.40×10-3 mol/L·atm.

Answer 1

0.02405 g/L is the solubility of argon in water at 25 °C.

Step-by-step explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

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

`C_(Ar)=K_H* p_(gas)`

where,

`K_H` = Henry's constant =
`1.40* 10^(-3)mol/L.atm`

`p_(Ar)` = partial pressure of carbonated drink = 0.51atm

Putting values in above equation, we get:

`C_(Ar)=1.40* 10^(-3)mol/L.atm* 0.430 atm\\\\C_(Ar)=6.02* 10^(-4)mol/L`

Molar mass of argon = 39.95 g/mol

Solubility of the argon gas :

`6.02* 10^(-4)mol/L* 39.95 g/mol=0.02405 g/L`

0.02405 g/L is the solubility of argon in water at 25 °C.