Question

What partial pressure of hydrogen gas is required in order for 0.00100 g of the gas to dissolve in 17.6 ml of pure water? the henry's law constant for hydrogen gas is 7.8 × 10–4 m atm–1?

Answer 1

To find the required partial pressure of hydrogen gas for its dissolution in water, convert the given mass of hydrogen to moles, calculate its molarity, and apply Henry's Law by dividing the molarity by the Henry's Law constant.

Step-by-step explanation:

Calculating Partial Pressure of Hydrogen in Water

To find the partial pressure of hydrogen gas required for the dissolution of 0.00100 g of H2 in 17.6 ml of water, we will use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The equation can be represented as: S = kH * P, where S is the solubility of the gas (in moles per liter), kH is the Henry's Law constant, and P is the partial pressure of the gas.

First, we convert the mass of H2 to moles using its molar mass (2.02 g/mol). Then, we calculate the molarity by dividing the number of moles by the volume of water in liters. Once we have the molarity, we can rearrange Henry's Law to solve for P: P = S / kH. By substituting the calculated molarity and the given Henry's Law constant for hydrogen, we can determine the required partial pressure.

Step-by-Step Calculation:

1. Convert mass of H2 to moles: moles = mass / molar mass

2. Calculate molarity: molarity = moles / volume in liters

3. Use Henry's Law to find partial pressure: P = S / kH

Answer 2

There is a minor typo in this question. The value of 7.8x10^-4 m/atm is nonsense. However, looking up the Henry's constant for hydrogen gas, there is a value of 7.8x10^-4 m/(L*atm) which I will assume is the correct unit and the rest of the problem will be done with that in mind.
First, determine what the molar concentration would be for 0.00100 g of H2 dissolved in 17.6 ml of water. Start with the atomic weight of hydrogen = 1.00794 g/mol. Molar mass of H2 is twice that, so 2 * 1.00794 = 2.01588 g/mol. So the number of moles of H2 we have is 0.001 g / 2.01588 g/mol = 0.000496061 mol. Finally, the molarity of the solution is 0.000496061 mol / 0.0176 L = 0.0281853 mol/L.
Now we can use the equation
Hcp = Ca/p
where
Hcp = Henry's constant
Ca = Concentration in aqueous solution
p = pressure
So solve for p, substitute the known values, and calculate:
Hcp = Ca/p
p*Hcp = Ca
p = Ca/Hcp
p = (0.0281853 mol/L)/(7.8x10^-4 mol/(L*atm))
p = 36.135 atm
Rounding to 3 significant figures gives p = 36.1 atmospheres.