Mole fraction of O2 in water = 2.73 x 10^-4 moles/L
what should I do to get the answer below:
= 4.194 x 10^-6 mol O2/mol H2O ?
(mass of water = 57208.99 kg/h)
If you have 2.73 x 10^-4 moles O2/L of solution, it means you essentially have 1L of water. That is, the amount of O2 is so small you can assume volume of solution = volume of water.
So if you have 1L of water at a density of 1.0g/mL you have 1000g of water. Using molar mass of water (18.0g/mol) you can calculate moles of water (=100g*1mol/18.0g).
Since you already have mol of O2 and you just calculate mol H2O you can now get the mol:mol ratio.
First, O2 mol fraction is not mol/L, mol fraction is mol O2/(mol O2+H2O). mol/L is volumetric molar concentration. Due the dilution, Jonathan and Clement have reason: you can assume that solution density is equal to water (1000 g/L at 4oC). But it is important that you understand that if you have greater volumetric molar concentration you need the density of solution for the conversion. Second, if you would have mol fraction “y=mol O2/(mol O2+H2O)” the conversion to molar relation “Y=mol O2/mol H2O” is direct: Y=y/(1-y).
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