Calculate the aqueous solubility (in milligrams per liter) of the following. The temperature is 20.°C in each case, the pressures are partial pressures of the gases.

(a) air at 2.4 atm
(b) He at 1.7 atm
(c) He at 11 kPa
There's also a table giving Henry's Constants for air and helium and they are 7.9 x 10^-4 and 3.7 x 10^-4 respectively.

Oh and the given molar mass of air was given to be 28.97g/mol

I looked here on the internet to see what form of Henry's Law you were/are using. Here is a link.

http://en.wikipedia.org/wiki/Henry%27s_law

It appears you are using pk = c with p in atm, and c in moles/L.
a)So pk = c
p = 2.4 atm and k = 7.9E-4
Substitute and solve for c. I find about 0.00190 M or 0.00190 moles/L.
moles = grams/molar mass; therefore, grams air = moles x molar mass air and that converted to mg should do it.
The others are done the same way.

To calculate the aqueous solubility of a gas, we can use Henry's Law, which states that the solubility of a gas is directly proportional to its partial pressure.

The equation for Henry's Law is:
S = k * P

Where:
S is the solubility of the gas in mol/L or mg/L,
k is the Henry's constant for the specific gas, and
P is the partial pressure of the gas in atmospheres (atm) or kilopascals (kPa).

In this case, we are given the partial pressures of air and helium and the Henry's constants for air and helium.

(a) To calculate the aqueous solubility of air at 2.4 atm:
Using Henry's Law equation, we have:
S = k * P
S = (7.9 x 10^-4) * 2.4
S ≈ 1.896 x 10^-3 mol/L

To convert solubility from mol/L to mg/L, you can multiply it by the molar mass of air, which is approximately 28.97 g/mol.

So, the solubility of air at 2.4 atm is:
S ≈ 1.896 x 10^-3 mol/L * 28.97 g/mol * 1000 mg/g
S ≈ 54.9 mg/L

Therefore, the aqueous solubility of air at 2.4 atm is approximately 54.9 mg/L.

(b) To calculate the aqueous solubility of helium at 1.7 atm:
Using Henry's Law equation, we have:
S = k * P
S = (3.7 x 10^-4) * 1.7
S ≈ 6.29 x 10^-4 mol/L

To convert solubility from mol/L to mg/L, you can multiply it by the molar mass of helium, which is approximately 4.00 g/mol.

So, the solubility of helium at 1.7 atm is:
S ≈ 6.29 x 10^-4 mol/L * 4.00 g/mol * 1000 mg/g
S ≈ 2.516 mg/L

Therefore, the aqueous solubility of helium at 1.7 atm is approximately 2.516 mg/L.

(c) To calculate the aqueous solubility of helium at 11 kPa:
Since the given Henry's constant for helium is in atm, we need to convert kilopascals to atmospheres.
1 atm = 101.325 kPa

Therefore, 11 kPa is approximately 11 / 101.325 ≈ 0.1084 atm.

Using Henry's Law equation, we have:
S = k * P
S = (3.7 x 10^-4) * 0.1084
S ≈ 4.006 x 10^-5 mol/L

To convert solubility from mol/L to mg/L, you can multiply it by the molar mass of helium, which is approximately 4.00 g/mol.

So, the solubility of helium at 11 kPa is:
S ≈ 4.006 x 10^-5 mol/L * 4.00 g/mol * 1000 mg/g
S ≈ 0.1602 mg/L

Therefore, the aqueous solubility of helium at 11 kPa is approximately 0.1602 mg/L.