An aqueous solution of 12.00% ammonium chloride, 99.5% Ammonium Chloride NH4Cl, by mass has a density of 1.036 g/mL. This means that in 1 L of the solution, we have 12.00%*1000= 120 g of NH4Cl.
To calculate the molality, we need to know the moles of NH4Cl present in 1 kg of solvent, which in this case is water. The molar mass of NH4Cl is 53.49 g/mol. Therefore, the number of moles of NH4Cl in 120 g is 120/53.49= 2.24 mol. The mass of 1 L of solution is 1.036 kg, and the mass of NH4Cl is 120 g. Thus, the mass of water is 1.036-0.120= 0.916 kg. The molality is then calculated as:
molality = moles of solute/ mass of solvent in kg
molality = 2.24 mol/ 0.916 kg
molality= 2.44 m
The mole fraction is the ratio of the number of moles of NH4Cl to the total number of moles of the solution. For our 1 L of solution, the molality is 2.24 mol. Using the density of the solution, we can calculate the total number of moles of the solution:
mass of solution= volume of solution * density of solution
mass of solution= 1 L * 1.036 g/mL= 1.036 kg
mass of water= mass of solution- mass of NH4Cl= 1.036-0.120= 0.916 kg
moles of water= mass of water/ molar mass of water= 0.916 kg/ 18.02 g/mol= 50.91 mol
Total moles of solution= moles of NH4Cl+ moles of water= 2.24 mol+ 50.91 mol= 53.15 mol
Therefore, the mole fraction of NH4Cl is:
mole fraction= moles of NH4Cl/ total moles of solution
mole fraction= 2.24 mol/ 53.15 mol= 0.0421
Finally, the molarity is the number of moles of NH4Cl in 1 L of solution. We have already calculated the moles of NH4Cl as 2.24 mol. Therefore, the molarity is:
molarity= moles of NH4Cl/ volume of solution in L
molarity= 2.24 mol/ 1 L
molarity= 2.24 M
In conclusion, a 12.00% ammonium chloride solution, 99.5% Ammonium Chloride NH4Cl, by mass has a molality of 2.44 m, a mole fraction of 0.0421, and a molarity of 2.24 M. These values can help in the preparation and analysis of the solution for various applications.