Question

The solubility of a gas in water is 0.66 g/L at 15 kPa of pressure. What is the solubility when the pressure is increased to 40.0 kPa?

Answer 1

The correct answer to the following question will be "1.76 g/L".

Step-by-step explanation:

The given values are:

P₁ = 15 kPa

Solubility of water, C₁ = 0.66 g/L

P₂ = 40.0 kPa

C₂ = ?

Now,

According to Henry's Law,

⇒
`(P_(1))/(P_(2))=(C_(1))/(C_(2))`

⇒
`C_(2)=(C_(1)* P_(2))/(P_(1))`

On putting the estimated values, we get

⇒
`=(0.66* 40)/(15)`

⇒
`=(26.4)/(15)`

⇒
`=1.76 \ g/L`