Exposing a 250 ml sample of water at 20.∘c to an atmosphere containing a gaseous solute at 27.59 kpa resulted in the dissolution of 4.51×10−3 g of the solute. use henry's law to determine the solubility of this gaseous solute when its pressure is 79.39 kpa.

Answer : The solubility of this gaseous solute will be,

Explanation :

First we have to calculate the concentration of solute.

Now we have to calculate the Henry's law constant.

Using Henry's law :

where,

C = concentration of solute =

p = partial pressure = 27.59 kPa

= Henry's law constant = ?

Now put all the given values in the above formula, we get:

Now we have to calculate the solubility of this gaseous solute when its pressure is 79.39 kPa.

Therefore, the solubility of this gaseous solute will be,

i believe it is c.) 1575 j.