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 1.600 atm and 20 oC. Hint, first find Henry's law constant at 20 oC and 1.00 atm.

I worked this for you yesterday.

To solve this problem, we need to use Henry's law, which states that the concentration (in this case, molal solubility) of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Step 1: Find Henry's Law constant at 20 oC and 1.00 atm:
Henry's Law constant (kH) can be calculated using the relation:
kH = (concentration of gas in the liquid) / (partial pressure of the gas)

Given:
Enthalpy of solution (ΔHsolution) = -12 kJ/mol
Solubility at 20 oC and 1.00 atm (S) = 0.121 g/100. g of water

Since we need to find kH, we need to rearrange the formula and convert the units:
ΔHsolution = kH * (partial pressure of the gas)
ΔHsolution in J/mol = kH * (partial pressure of the gas in atm * atmospheric pressure in Pascals)

Converting units:
ΔHsolution = -12 * 1000 J/mol (since 1 kJ = 1000 J)
Partial pressure of the gas = 1.00 atm
Atmospheric pressure at 20 oC = 1.00 atm * 101325 Pa/atm = 101325 Pa

Now, rearrange the equation to find kH:
kH = ΔHsolution / (partial pressure of the gas * atmospheric pressure)
kH = (-12 * 1000 J/mol) / (1.00 atm * 101325 Pa)

kH = -118.57 J/(mol * Pa)

Note: In this case, kH will have units of J/(mol * Pa) instead of the commonly used units of mol/(L * atm).

Step 2: Calculate molal solubility at 1.600 atm:
Given:
Partial pressure of the gas (P) = 1.600 atm

Using Henry's Law, we can write:
kH = S / P

Rearranging the equation to solve for S:
S = kH * P

Substituting the given values:
S = -118.57 J/(mol * Pa) * 1.600 atm

Since we need to find the molal solubility, we need to convert the units:
1 atm = 101325 Pa
Therefore, 1.600 atm = 1.600 * 101325 Pa

S = -118.57 J/(mol * Pa) * (1.600 * 101325 Pa)

Solving the equation:
S = -19,049 J/mol

To convert this to molal solubility, we need to convert J to kJ and divide by the molecular weight to get molality (mol/kg).

Given:
Molecular weight of nitrous oxide (N2O) = 44.013 g/mol

S(molality) = (-19,049 J/mol) / (44.013 g/mol)
S(molality) = -432.7 mol/(kg * J)

Note: The answer will have units of mol/(kg * J) instead of the commonly used units of mol/kg.

So, the molal solubility of nitrous oxide in water at 1.600 atm and 20 oC is approximately -432.7 mol/(kg * J).