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This problem is providing us with the initial solubility and pressure of a gas and asks for the solubility once the pressure is changed to 10 kPa. At the end, the answer is found to be 0.4 g/L.
In chemistry, the concept of ideal gas provides us first approach to the understanding of gases, as a model with no interactions and perfectly spherical-shaped molecules, however, this is not an actual a fact in nature.
Thus, we can relate pressure, volume, temperature and moles (or grams) by using the widely-known ideal gas equation:
PV = nRT
Which in the case of the solubility of the gas, can be written as:
P=Sol*RT*MM
Which at constant temperature can be just:
[tex]\frac{P}{Sol} =k[/tex]
Since R, T and MM (molar mass of the gas) are constants, and for two states:
[tex]\frac{P_1}{Sol_1} =\frac{P_2}{Sol_2}[/tex]
In such a way, one can calculate the solubility when the pressure changes as follows:
[tex]Sol_2=\frac{Sol_1*P_2}{P_1}[/tex]
Finally, we plug in the numbers to obtain:
[tex]Sol_2=\frac{2g/L*10kPa}{50kPa} =0.4g/L[/tex]
Learn more about ideal gases: brainly.com/question/8711877