Question

The solubility of nitrogen gas at 25 degrees C and 1 atm is 6.8 x 10^(-4) mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm, what is the concentration (molarity) of dissolved nitrogen?

5.2 x 10^(-4) M
1.1 x 10^(-5) M
4.9 x 10^(-4) M
3.8 x 10^(-4) M
6.8 x 10^(-4) M

Answer 1

Answer:

5.2 x 10⁻⁴ M.

Explanation:

• The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

P = kC

where P is the partial pressure of the gaseous  solute above the solution.

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas.

• At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:

P₁C₂ = P₂C₁,

P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.

P₂ = 0.76 atm, C₂ = ??? mol/L.

∴ C₂ = (P₂C₁)/P₁ = (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = 5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.

Answer 2

5.2 × 10⁻⁴ M is the concentration of dissolved nitrogen, if the partial pressure of nitrogen gas in air is 0.76 atm.

What is Henry's Law?

Henry's law is proposed for the dissolved gas and it is represented as:

P = kC, where

P = partial pressure

k = henry's constant

C = concentration

And for the given question required equation for concentration is :

P₁/C₁ = P₂/C₂

P₁C₂ = P₂C₁, where

P₁ = given partial pressure of nitrogen = 1atm

C₁ = given concentration of nitrogen = 6.8×10⁻⁴ mole/L

P₂ = partial pressure of nitrogen = 0.76atm

C₂ = required concentration of nitrogen = to find?

On putting all values in the above equation we get,

1atm × C₂ = 0.76atm × 6.8×10⁻⁴ mole/L

C₂ = 5.168 x 10⁻⁴ mol/L = 5.2 × 10⁻⁴ M

Hence, option (1) is correct i.e. 5.2 × 10⁻⁴ M is the concentration of dissolved nitrogen.

To know more about henry's law, visit the below link:

https://brainly.com/question/7007748