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The solubility of argon in water at 25 degree C is 0.150 mol/L. What is the Henry's Law constant for argon if the partial pressure of argon in air is 0.0984 atm? a. 1.52 mole/L middot atm b. 0.0164 mole/L middot atm c. 12.5 mole/L middot atm d. 0.656 mole/L middot atm e. 0.0801 mole/L middot atm

Respuesta :

The Henry's Law constant for argon if the partial pressure of argon in air is 0.0984 atm is 1.52 mole/L *atm.

Henry's regulation states that at a regular temperature, the amount of a given fuel that dissolves in a liquid is immediately proportional to the partial pressure of that gas in equilibrium with that liquid. Henry's regulation states that the solubility of a fuel in a liquid is immediately proportional to the pressure of the gas.

Pressure of argon =  p = 0.0984 atm

Molar solubility at 25°C, S = 0.150 mol/l

Calculating the Henry's Law constant,

     S =KP

       K =S/P

         0.150/0.0984

         = 1.52 mol/L atm

Hence the  Henry's Law constant (K) = = 1.52 mol/L atm

Learn more about Henry's Law here:-https://brainly.com/question/23204201

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