Commercial concentrated aqueous ammonia is 28\(\% \mathrm{NH}_{3}\) by mass and has a density of 0.90 \(\mathrm{g} / \mathrm{mL} .\) What is the molarity of this solution?

Short Answer

Expert verified
The molarity of the commercial concentrated aqueous ammonia solution is approximately 14.8 M.

Step by step solution

01

Formula for molarity

Molarity (M) is defined as the number of moles of solute (in this case, NH3) present in 1 liter of the solution. The formula for molarity is: \[M = \frac{moles \ of \ solute}{volume \ of \ solution \ (L)}\] #Step 2: Calculate the mass of NH3 in 1 mL of the solution#
02

Mass of NH3 in 1 mL

We are given that the concentrated ammonia solution is 28% NH3 by mass. Therefore, the mass of NH3 in 1 mL solution can be calculated as: Mass of NH3 in 1 mL = (percentage composition) × (mass of 1 mL solution) We are given that the density of the solution is 0.90 g/mL, so the mass of 1 mL solution = 0.90 g Mass of NH3 = 0.28 × 0.90 g = 0.252 g #Step 3: Calculate the moles of NH3 in 1 mL of the solution#
03

Moles of NH3 in 1 mL

We are given the mass of NH3 in 1 mL of the solution as 0.252 g. To convert this to moles, we will need to use the molecular weight of NH3. NH3 has one nitrogen atom (14 g/mol), and three hydrogen atoms (3 × 1 g/mol), so the molecular weight of NH3 is (14 + 3) g/mol = 17 g/mol. The moles of NH3 in 1 mL of the solution is: Moles of NH3 = mass of NH3 ÷ molar mass of NH3 Moles of NH3 = 0.252 g ÷ 17 g/mol ≈ 0.0148 mol #Step 4: Calculate the molarity of NH3 in the solution#
04

Molarity of NH3

Now that we know the moles of NH3 in 1 mL of the solution, we can convert this to the molarity of NH3 in the solution. Recall the formula for molarity: \[M = \frac{moles \ of \ solute}{volume \ of \ solution \ (L)}\] To find the molarity of the solution, we can divide the moles of NH3 in 1 mL of the solution by the volume of 1 mL in liters (0.001 L): Molarity = 0.0148 mol ÷ 0.001 L ≈ 14.8 M #Solution# The molarity of the commercial concentrated aqueous ammonia solution is approximately 14.8 M.

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