The solubility of argon in water at 25°C is 0.0150 mol/L. What is the Henry's Law constant for argon if the partial pressure of argon in air is 0.00934 atm?
0.61mol/L
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To find the Henry's Law constant for argon, we need to use the equation that relates the solubility of a gas to its partial pressure. The equation is as follows:
S = k * P
Where:
- S is the solubility of the gas in moles per liter (mol/L)
- k is the Henry's Law constant
- P is the partial pressure of the gas in atmospheres (atm)
We are given the solubility of argon in water at 25°C, which is 0.0150 mol/L. We are also given the partial pressure of argon in air, which is 0.00934 atm.
Now we can rearrange the equation to solve for the Henry's Law constant:
k = S / P
Substituting the given values:
k = 0.0150 mol/L / 0.00934 atm
Calculating the value:
k ≈ 1.61 mol/(L·atm)
Therefore, the Henry's Law constant for argon at 25°C is approximately 1.61 mol/(L·atm).
http://www.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Henry%27s_law.html
k=p/c so the number is dependent on the units you choose. If p is i atm, and c in mol/L. In some texts, K is defined differently.
k=.00934/.0150 atm*L/mol