Question

What partial pressure of nitrogen gas is required in order for 0.00134 g of the gas to dissolve in 13.1 ml of pure water? the henry's law constant for nitrogen gas is 6.1 ´ 10⁻⁴ m atm⁻¹.

a) 6.2 ´ 10⁻¹ atm
b) 1.7 ´ 10⁻¹ atm
c) 6.0 ´ 100 atm
d) 2.9 ´ 10⁻⁸ atm
e) 1.7 ´ 10⁻² atm

Answer 1

To determine the partial pressure of nitrogen gas required for it to dissolve in water, we can use Henry's law equation which states that the solubility of a gas is directly proportional to its partial pressure.

Step-by-step explanation:

To determine the partial pressure of nitrogen gas required for it to dissolve in water, we can use Henry's law equation which states that the solubility of a gas is directly proportional to its partial pressure.

Henry's law equation is:

C = k * P

where C is the concentration of the gas in the solution, k is the Henry's law constant, and P is the partial pressure of the gas.

In this case, we are given the mass of nitrogen gas (0.00134 g) and the volume of water (13.1 mL).

We can first convert the mass of nitrogen gas to moles using its molar mass (28.01 g/mol).

moles = mass / molar mass = 0.00134 g / 28.01 g/mol = 4.789 × 10^-5 mol

Next, we can convert the volume of water to liters:

volume = 13.1 mL / 1000 mL/L = 0.0131 L

Now we can substitute these values into Henry's law equation:

C = k * P

4.789 × 10^-5 mol / 0.0131 L = (6.1 × 10^-4 mol/(L·atm)) * P

Solving for P:

P = (4.789 × 10^-5 mol / 0.0131 L) / (6.1 × 10^-4 mol/(L·atm))

P ≈ 6.2 × 10^-1 atm

Therefore, the partial pressure of nitrogen gas required for it to dissolve in water is approximately 6.2 × 10^-1 atm.