The enthalpy of solution of nitrous oxide (N2O) in water is - 12 kJ/mol and its solubility at 20 oC and 1.00 atm is 0.121 g per 100. g of water. Calculate the molal solubility of nitrous oxide in water at 1.400 atm and 20 oC. Hint, first find Henry's law constant at 20 oC and 1.00 atm.

0.42 m

Explanation:

First, we need to know which is the Henry's law. Henry's law states the following:

"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid".

In summary, is the following expression:

C = k*Pi (1)

Where:

C: Solubility of the gas (m)

Pi: Partial pressure of the gas (atm)

k: constant of henry's law (in m/atm)

Now with the first data of pressure and solubility we can calculate the constant k, however, the solubility must be in mol/kg, and we have solubility in g/gH2O. Let's turn this solubility in mol, using the molar mass of N2O, which is 44 g/mol:

moles N2O = 0.121 / 44 = 0.003 moles

molality = moles / kg water

m = 0.003 / 0.1 = 0.03 moles/kg

Now that we have the solubility in m, we can calculate the constant:

k = C/Pi

k = 0.03/1 = 0.03 m/atm

With this value, we only need to apply Henry's law, with the new pressure of 1.4 atm and solve for C:

C = 0.03 * 1.4

C = 0.42 moles/kg water