We're asked to find the mass percentage and the molality of a solution, given the masses of solute and solvent.
The percentage by mass of the solute is given by
#%"mass solute" = "g solute"/"g solution" xx 100%#
We know:
The mass of the solution is the combined mass of solute and solvent:
#"g solution" = 84.6# #"g solute"# #+ 822# #"g solvent"# #= color(red)(907# #color(red)("g solution"#
We therefore have
#%"mass NH"_3 = (84.6color(white)(l)"g NH"_3)/(color(red)(907color(white)(l)"g solution")) xx 100% = color(blue)(9.33%#
The molality of a solution is
#"molality" = "mol solute"/"kg solvent"#
Convert from grams of #"NH"_3# to moles using its molar mass (#17.03# #"g/mol"#):
#84.6cancel("g NH"_3)((1color(white)(l)"mol NH"_3)/(17.03cancel("g NH"_3))) = 4.97# #"mol NH"_3#
The kilograms of solvent (water) is
#822cancel("g H"_2"O")((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = 0.822# #"kg H"_2"O"#
The molality is thus
#"molality" = (4.97color(white)(l)"mol NH"_3)/(0.822color(white)(l)"kg H"_2"O") = color(green)(6.04m#
