the solubility of argon in water at 25 degrees celcius is 0.0150M. what is the henry's law constant of Ar if the partial pressure of argon is 0.00934 atm.
A)0.000140 M/atm
B)0.623 M/ atm
C)1.61 M/atm
D) 4.10 M/atm
K=Pgas/Cgas
Post your work if you get stuck.
1.61
To determine the Henry's law constant (K) for argon (Ar), we can use the formula:
K = Pgas / Cgas
Given:
Partial pressure of argon (Pgas) = 0.00934 atm
Solubility of argon in water (Cgas) = 0.0150 M
Substituting the given values into the formula:
K = 0.00934 atm / 0.0150 M
Calculating the result:
K = 0.623
Therefore, the Henry's law constant for argon (Ar) is 0.623 M/atm.
The correct answer is B) 0.623 M/atm.
To solve this problem, we can use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The Henry's law constant (K) relates the solubility (C) to the partial pressure (P) and is given by the equation K = P/C.
In this case, we are given the solubility of argon in water at 25 degrees Celsius (0.0150 M) and the partial pressure of argon (0.00934 atm). We need to find the Henry's law constant of argon.
Substituting the given values into the equation K = P/C, we have:
K = (0.00934 atm) / (0.0150 M)
To calculate this, we need to convert the concentration from moles per liter (M) to moles per liter per atmosphere (M/atm). The conversion factor is 1 M = 1 L/atm.
K = (0.00934 atm) / (0.0150 M) * (1 L/atm) = 0.6227 M/atm
The calculated value of K is approximately 0.6227 M/atm.
Comparing the calculated value to the given options, we can see that the closest option is:
B) 0.623 M/atm
Therefore, the correct answer is B) 0.623 M/atm.