For a solution with benzene as the solvent, what is the molality (with respect to the solute) if the solution has a solute mol fraction of #0.2#?
1 Answer
Apr 19, 2017
Since the mol fraction of solute is
#chi_i = 0.2 = (n_i)/(n_i+n_j)#
#= ("0.2 mols solute")/("0.2 mols solute" + "0.8 mols solvent")# where
#n_i# is the mols of#i# .
Recall that molality is:
#"mols solute"/"kg solvent"#
Fortunately, we know that the solvent is benzene, so the molar mass is:
#M_(j) = 6 xx 12.011 "g/mol C" + 6 xx 1.0079 "g/mol H"#
#=# #"78.1134 g/mol"#
This tells us that the molality is:
#color(blue)(m_"soln") = "0.2 mols solute"/(0.8 cancel"mols benzene") xx (cancel"1 mol benzene")/(78.1134 cancel"g benzene") xx (1000 cancel"g")/"1 kg"#
#=# #color(blue)("3.20 molal")#