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  • At 298 K, the solubility of oxygen gas in water is . <p dir="lt

At 298 K, the solubility of oxygen gas in water is 7.91\times\10^{-4}M.

The concentration of oxygen gas in equilibrium with the air is 0.21 atm. What is the Henry's law constant for oxygen gas in water?

Option: 1

1.19\times\10{^3} M/atm


Option: 2

2.39\times\10{^3} M/atm


Option: 3

3.58\times\10{^3} M/atm


Option: 4

4.77\times\10{^3} M/atm


Answers (1)

best_answer

 Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Hence, we can use the equation: C = k_H \times P

Where C is the concentration of the gas in the liquid, k_H is the Henry's Law constant, and P is the partial pressure of the gas above the liquid.

Rearranging this equation to solve for k_H, we get:

k_H = \frac{C}{P}

Given that the concentration of oxygen gas in equilibrium with air is 0.21 atm and the solubility of oxygen gas in water is 7.91\times 10^{-4} M, we have:

k_H = 7.91\times \10^{-4} M / 0.21 atm

k_H = 2.3\times 10{^3}M/atm

 

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shivangi.bhatnagar

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