Question

A student dissolves of ammonia in of a solvent with a density of . The student notices that the volume of the solvent does not change when the ammonia dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer 1

Answer:

1.2 M

1.2 m

Explanation:

There is some info missing. I think this is the original question.

A student dissolves 5.1g of ammonia (NH₃) in 250 mL of a solvent with a density of 1.02 g/mL. The student notices that the volume of the solvent does not change when the ammonia dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.

Molarity

The molar mass of ammonia is 17.03 g/mol. The moles corresponding to 5.1 grams is:

5.1 g × (1 mol/17.03 g) = 0.30 mol

The volume of the solution is equal to the volume of the solvent, 250 mL or 0,250 L.

The molarity of ammonia is:

M = moles of solute / liters of solution

M = 0.30 mol / 0.250 L

M = 1.2 M

Molality

The density of the solvent is 1.02 g/mL. The mass corresponding to 250 mL is:

250 mL × 1.02 g/mL = 255 g = 0.255 kg

The molality of ammonia is:

m = moles of solute / kilograms of solvent

m = 0.30 mol / 0.255 kg

m = 1.17 m ≈ 1.2 m