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#