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
