If the solubility of O₂ at 0.360 bar and 25 $\circ$ C is 15.0 g/100 g H₂O,what is the solubility of O₂ at a pressure of 1.72 bar and 25 $\circ$ C? A) 71.7 g/100 g H₂O B) 24.2 g/100 g H₂O C) 0.319 g/100 g H₂O D) 0.0140 g/100 g H₂O E) 3.14 g/100 g H₂O

Question

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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. Mathematically, this can be expressed as:S = kPwhere S is the solubility of the gas in g/100 g H2O, k is the Henry's law constant, and P is the partial pressure of the gas in bar.In this problem, we are given the solubility of O2 at 0.360 bar and 25°C as 15.0 g/100 g H2O. We can use this information to find the Henry's law constant for O2:k = S/P = 15.0 g/100 g H2O / 0.360 bar = 41.7 g/100 g H2O barNow we can use the Henry's law constant to find the solubility of O2 at a pressure of 1.72 bar and 25°C:S = kP = 41.7 g/100 g H2O bar * 1.72 bar = 71.7 g/100 g H2OTherefore, the solubility of O2 at a pressure of 1.72 bar and 25°C is 71.7 g/100 g H2O.

If the Solubility of O2 at 0 ∘\circ∘ C Is 15 ∘\circ∘

If the Solubility of O₂ at 0 $\circ$ C Is 15 $\circ$