If a gaseous sample contains 74% N2 by volume, what is the solubility of N2 in water at 25°C and 1.0 atm (kH in H2O at 25°C = 7.0 10-4 mol/Latm)?
C = K*p. The mol fraction N2 is 0.74 and partial pressure (pN2) = 0.74 x total P or 0.74 x 1 = 0.74 atm
Then C = 7.0E-4 x 0.74 = ?M
To find the solubility of N2 in water at 25°C and 1.0 atm, we can use Henry's Law. Henry's Law states that the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.
The equation for Henry's Law is:
C = kH * P
Where:
C is the concentration of the gas in the liquid (in this case, N2 in water)
kH is the Henry's Law constant for the gas in the specific liquid (given as 7.0 * 10^(-4) mol/L*atm)
P is the partial pressure of the gas (1.0 atm)
To find the concentration of N2 in water, we need to convert the percentage by volume to the partial pressure of N2 in the gas. Since the gaseous sample contains 74% N2 by volume, we can consider that the partial pressure of N2 is 74% of the total pressure of the gas sample.
Let's assume the total pressure of the gas sample is 1.0 atm (as given). Therefore, the partial pressure of N2 would be:
Partial pressure of N2 = 74% * 1.0 atm
= 0.74 atm
Now, we can substitute the values into the Henry's Law equation:
C = kH * P
= (7.0 * 10^(-4) mol/L*atm) * (0.74 atm)
Calculate the product:
C = 5.18 * 10^(-4) mol/L
So, the solubility of N2 in water at 25°C and 1.0 atm is approximately 5.18 * 10^(-4) mol/L.