Question

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

The molarity of the commercial concentrated aqueous ammonia solution is 14.82 M.

Step-by-step solution

Calculate the mass of ammonia in 1L of solution

: First, we need to find out the mass of ammonia (NH3) in 1L of the solution. Given that the density of the solution is 0.90 g/mL, we can determine the mass of 1L of the solution by multiplying the density by the volume: Mass of 1L solution = Density × Volume Mass of 1L solution = \(0.90 \frac{\mathrm{g}}{\mathrm{mL}} × 1000 \mathrm{mL}\) Mass of 1L solution = 900 g Now, we can calculate the mass of ammonia in 1L of the solution by using the given percentage by mass: Mass of ammonia = (Percentage by mass × Mass of 1L solution) / 100 Mass of ammonia = \(28\% × 900 \mathrm{g}\) / 100 Mass of ammonia = 252 g

Convert the mass of ammonia to moles

: Next, we need to convert the mass of ammonia to moles using its molar mass. The molar mass of NH3 is approximately 17 g/mol (14 g/mol for N and 3 g/mol for H). Moles of ammonia = Mass of ammonia / Molar mass of ammonia Moles of ammonia = \(252 \mathrm{g} / 17 \frac{\mathrm{g}}{\mathrm{mol}}\) Moles of ammonia = 14.82 moles

Calculate the molarity of the solution

: Finally, we can calculate the molarity of the ammonia solution using the moles of ammonia and the volume of the solution. Molarity = Moles of solute / Volume of solution (in Liters) Molarity = \(14.82 \mathrm{moles} / 1 \mathrm{L}\) Molarity = 14.82 M Therefore, the molarity of the commercial concentrated aqueous ammonia solution is 14.82 M.

Key concepts

Aqueous Solutions

An aqueous solution is simply a mixture where water acts as the solvent. The term "aqueous" indicates that the substance is dissolved in water, and the solute is typically some other substance, like ammonia in this exercise. Aqueous solutions are crucial in chemistry because they easily dissolve ionic compounds, enabling chemical reactions. For instance:

• They play a fundamental role in biochemical processes in our bodies.
• They are used extensively in laboratories to conduct experiments.
• Many household and industrial products are manufactured using aqueous solutions.

Understanding how substances behave in water allows us to predict reactions and calculate concentrations, such as molarity, effectively.

Density Formula

The density formula is a key concept when working with solutions. It is defined as the mass of an object divided by its volume:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]In this exercise, we understand that density helps us find the total mass of the solution when we know its volume. For example, with a density of \(0.90 \text{ g/mL}\), multiplying by 1000 mL gives you the mass of 1 liter of solution, which is 900 g. Using density in calculations helps:

• Determine how concentrated a solution is.
• Convert between volumes and masses, essential in making solutions.

Grasping density enables accurate preparation and analysis of solutions.

Percentage by Mass

Percentage by mass is an approach to express the concentration of a component in a solution. It is calculated as:\[ \text{Percentage by mass} = \left(\frac{\text{Mass of solute}}{\text{Total mass of solution}} \right) \times 100\]In the given problem, ammonia makes up 28% of the solution by mass. This means:

• Out of 900 g of solution, 252 g is ammonia.
• It is useful in determining how much of a solute is present compared to the whole solution.

Using percentage by mass allows chemists to easily compare concentrations of different solutions.

Molar Mass

Molar mass is another crucial concept that links mass and moles, helping to calculate the number of moles from a given mass of a substance. It is the mass of one mole of a substance, expressed in g/mol. For ammonia (NH3), the molar mass is computed by summing the atomic masses of nitrogen (N) and hydrogen (H):\[ \text{Molar mass of NH}_3 = 14 + 3 \times 1 = 17 \text{ g/mol} \]Understanding molar mass is vital for converting between mass and moles, such as in the problem where 252 g of ammonia equals 14.82 moles. Highlights of using molar mass include:

• Facilitating stoichiometric calculations in chemical reactions.
• Allowing for correct conversion in calculating solution concentrations.

Mastering molar mass ensures accurate composition and concentration calculations in chemistry.