At 20 ° C the Henry's law constant for argon in water is 1.64×10^{- 3} molal/atm. How many grams of Ar (atomic mass = 39.9 g/mol) will dissolve in 500 g of water (molar mass = 18.0 g/mol) at 20 ° C if the pressure of Ar above the solution is 2.5 atm?

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Quizplus · Accepted Answer

To find the grams of Ar dissolved, we first calculate the moles of Ar using Henry's law, which states that the solubility of a gas in a liquid at a given temperature is directly proportional to the pressure of the gas above the liquid. The formula is $S = kH \cdot P$, where $S$ is the solubility in mol/kg, $kH$ is Henry's law constant, and $P$ is the pressure.Given:- $kH = 1.64 \times 10^{-3} \, \text{molal/atm}$- $P = 2.5 \, \text{atm}$- $500 \, \text{g water} = 500 \, \text{g} \times \frac{1 \, \text{mol}}{18.0 \, \text{g}} = 27.78 \, \text{mol water}$The solubility $S$ is calculated as:$$S = 1.64 \times 10^{-3} \times 2.5 = 4.1 \times 10^{-3} \, \text{mol/kg}$$To find the moles of Ar dissolved in 500 g (or 0.5 kg) of water:$$4.1 \times 10^{-3} \, \text{mol/kg} \times 0.5 \, \text{kg} = 2.05 \times 10^{-3} \, \text{mol Ar}$$To convert moles of Ar to grams:$$2.05 \times 10^{-3} \, \text{mol} \times 39.9 \, \text{g/mol} = 0.0818 \, \text{g}$$Therefore, 0.082 g of Ar will dissolve in 500 g of water under the given conditions.

At 20 ° C the Henry's Law Constant for Argon