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
1 answer:
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:
<em>P = kC </em>
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: <em>P₁C₂ = P₂C₁, </em>
P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.
P₂ = 0.76 atm, C₂ = ??? mol/L.
<em>∴ C₂ = (P₂C₁)/P₁ = </em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M. </em>
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