# Question d4d24

Jul 16, 2017

"mass percentage" = 9.33% ${\text{NH}}_{3}$

$\text{molality} = 6.04 m$

#### Explanation:

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:

• $84.6$ $\text{g solute}$ (ammonia)

• $822$ $\text{g solvent}$ (water)

The mass of the solution is the combined mass of solute and solvent:

$\text{g solution} = 84.6$ $\text{g solute}$ $+ 822$ $\text{g solvent}$ = 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

$\text{molality" = "mol solute"/"kg solvent}$

Convert from grams of ${\text{NH}}_{3}$ to moles using its molar mass ($17.03$ $\text{g/mol}$):

84.6cancel("g NH"_3)((1color(white)(l)"mol NH"_3)/(17.03cancel("g NH"_3))) = 4.97 ${\text{mol NH}}_{3}$

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 $\text{kg H"_2"O}$

The molality is thus

"molality" = (4.97color(white)(l)"mol NH"_3)/(0.822color(white)(l)"kg H"_2"O") = color(green)(6.04m#