Question

What is the molality, m, of an aqueous solution of ammonia that is 12.83 M NH3 (17.03 g/mol)? This solution has a density of 0.9102 g/mL.

Answer 1

Answer:

Molality = 18.5 m

Explanation:

Let's analyse data. We want to determine molality which means mol of solute / 1kg of solvent. (Hence we need, the moles of solute and the mass of solvent in kg)

12.83 M means molarity → mol of solute in 1L of solution

Density refers always to solution → Mass of solution / Volume of solution

1L = 1000 mL

We can determine the mass of solution with density

0.9102 g/mL = Mass of solution / 1000 mL

Mass of solution = 0.9102 g/mL . 1000 mL → 910.2 g

Let's convert the moles of solute (NH₃) to mass

12.83 mol . 17.03 g/ 1 mol = 218.5 g

We can apply this knowledge:

Mass of solution = Mass of solvent + Mass of solute

910.2 g = Mass of solvent + 218.5 g

910.2 g - 218.5 g = 691.7 g → Mass of solvent.

Let's convert the mass in g to kg

691.7 g . 1kg / 1000 g = 0.6917kg

We can determine molalilty now → 12.83 mol / 0.6917kg

Molality = 18.5 m