Question

The Henry's Law constant (kH) for carbon monoxide in water at 25°C is 9.71×10-4 mol/L·atm. How many grams of CO will dissolve in 1.00 L of water if the partial pressure of CO is 2.75 atm?

Answer 1

Answer: 0.0748 grams of CO will dissolve in 1.00 L of water if the partial pressure of CO is 2.75 atm

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_(CO)=K_H* p_(liquid)`

where,

`K_H` = Henry's constant =
`9.71* 10^(-4)mol/L.atm`

`p_(CO)` = partial pressure of CO = 2.75 atm

Putting values in above equation, we get:

`C_(CO)=9.71* 10^(-4)mol/L.atm* 2.75atm\\\\C_(CO)=2.67* 10^(-3)mol/L`

`C_(CO)=2.67* 10^(-3)mol/L* 28g/mol=0.0748g/L`

Hence, the solubility of carbon monoxide gas is 0.0748 g/L