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the questions says
at 0°C gas is dissolved in H2O at a mol fraction of 2 x 10^-18. What is the molality of the solution given the molecular weight of the gas is 32 g/mol.
so we know:
molality = mols of solute/kg of solvent.
so the explanation in the book is that you can assume that there is 1 mol of water, which weights 0.018 kg.
the explanation goes on to say that to find molality, it would m = (2 x 10^-18)/ 0.018 = 1 x 10^-16
i'm confused about why we can assume that there is 2 x 10^-18 of the solute when we also have to consider the mols of the water since mol fraction = mols of solute/ total mols
..so, technically, shouldn't the formula for mol fraction be... mols solute/ (mols of solute + 1 molH2O) = 2 x 10^-18
at 0°C gas is dissolved in H2O at a mol fraction of 2 x 10^-18. What is the molality of the solution given the molecular weight of the gas is 32 g/mol.
so we know:
molality = mols of solute/kg of solvent.
so the explanation in the book is that you can assume that there is 1 mol of water, which weights 0.018 kg.
the explanation goes on to say that to find molality, it would m = (2 x 10^-18)/ 0.018 = 1 x 10^-16
i'm confused about why we can assume that there is 2 x 10^-18 of the solute when we also have to consider the mols of the water since mol fraction = mols of solute/ total mols
..so, technically, shouldn't the formula for mol fraction be... mols solute/ (mols of solute + 1 molH2O) = 2 x 10^-18