How do I convert mole fraction to molality? Then also to percent by mass?
Convert 0.1538 mole fraction of propanol (C3H8O) in water to molality. 4sig figs. Then convert the same mole fraction to percent by mass.
Convert 0.1538 mole fraction of propanol (C3H8O) in water to molality. 4sig figs. Then convert the same mole fraction to percent by mass.
1 Answer
Well, you should assume a certain number of mols of solution.
#chi_("C"_3"H"_8"O") = 0.1538 = "mols solute"/"mols solution"#
#= ("0.1538 mols C"_3"H"_8"O")/("0.1538 mols C"_3"H"_8"O" + "0.8462 mols H"_2"O")#
As a result, the mass of water in
#0.8462 cancel("mols H"_2"O") xx ("18.015 g H"_2"O")/cancel("1 mol H"_2"O") = "15.24 g"#
#=# #"0.01524 kg water"#
Therefore, the molality is:
#color(blue)(m) = "mols solute"/"kg solvent"#
#= ("0.1538 mols C"_3"H"_8"O")/("0.01524 kg H"_2"O")#
#=# #color(blue)("10.09 mol/kg")#
The percent by mass might even be easier; all you have to do now, now that you have mols of each component, is convert the mols to grams within the mol fraction:
#color(blue)("%w/w") = (0.1538 cancel("mols C"_3"H"_8"O") xx ("60.096 g C"_3"H"_8"O")/cancel("1 mol"))/(0.1538 cancel("mols C"_3"H"_8"O") xx ("60.096 g C"_3"H"_8"O")/cancel("1 mol") + 0.8462 cancel("mols H"_2"O") xx ("18.015 g H"_2"O")/cancel("1 mol")) xx 100%#
#= color(blue)(37.75%)#
