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An aqueous solution contains \(4.00 \% \mathrm{NH}_{3}\) (ammonia) by mass. The density of the aqueous ammonia is \(0.979 \mathrm{~g} / \mathrm{mL}\). What is the molarity of \(\mathrm{NH}_{3}\) in the solution?

Short Answer

Expert verified
The molarity of the NH₃ solution is 2.30 M.

Step by step solution

01

Understanding the Percentage by Mass

The solution contains 4.00% NH₃ by mass, which means that in 100 g of solution, there are 4 g of NH₃ and 96 g of water.
02

Calculate the Total Mass of the Solution

For calculations, assume you have 1000 mL (1 liter) of solution. Using the given density of 0.979 g/mL, the mass of the solution is: \[ 1000 \text{ mL} \times 0.979 \frac{\text{g}}{\text{mL}} = 979 \text{ g} \]
03

Calculate the Mass of NH₃ in 1000 mL

With a 4.00% composition, the mass of NH₃ in 979 g of solution is: \[ 0.04 \times 979 \text{ g} = 39.16 \text{ g} \]
04

Convert Grams of NH₃ to Moles

The molar mass of NH₃ is 17.03 g/mol. Convert the mass of NH₃ to moles using its molar mass: \[ \frac{39.16 \text{ g}}{17.03 \frac{\text{g}}{\text{mol}}} = 2.30 \text{ moles of NH}_3 \]
05

Calculate Molarity

Molarity is defined as the number of moles of solute per liter of solution. Since we have calculated for 1 liter of solution: \[ \text{Molarity} = \frac{2.30 \text{ moles}}{1 \text{ L}} = 2.30 \text{ M} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aqueous Solution
An aqueous solution is a mixture where water acts as the solvent. A solvent is the substance that dissolves other substances, known as solutes, which in this case is ammonia (NH₃). Aqueous solutions are characterized by the ability to dissolve a great variety of substances, notably ionic compounds, due to water's polar nature.
An example of an aqueous solution is when you dissolve table salt in water. Similarly, the described solution contains ammonia dissolved in water. Understanding aqueous solutions is crucial for various scientific processes, as they are often used in chemical reactions, biological systems, and industrial applications.
In your calculations, water serves as the primary medium, making it essential to consider both its mass and volume when exploring the concentration of solutes, like NH₃.
Percentage by Mass
The percentage by mass, often referred to as mass percent, is a way to express the concentration of a component in a mixture. It is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100.
This concept is crucial because it gives a straightforward and universally understandable measure of concentration. In the provided solution, the percentage by mass of ammonia is 4.00%.
This implies that in 100 grams of the solution, 4 grams are ammonia, and the remaining 96 grams are water. Understanding this helps in determining how much of a particular substance is present within a given amount of solution.
Density of Solution
Density is a vital property in chemistry that relates the mass of a substance to its volume, given in units such as grams per milliliter (g/mL). It is an important concept when dealing with solutions, as it allows you to determine how much mass is within a certain volume.
In the original exercise, the density of the aqueous ammonia solution is 0.979 g/mL. This means for every milliliter of the solution, you have 0.979 grams of total mass.
Understanding density helps to transition from mass percentages or mass itself to volume-related measurements, which is often required in tasks like calculating molarity.
Moles of Solute
The mole is a fundamental concept in chemistry used to count entities at the atomic scale. One mole corresponds to Avogadro's number, which is approximately 6.022 × 10²³ entities. When referring to moles of solute, it means the number of moles of the substance dissolved in a solution.
To determine how many moles of ammonia are present in the solution, we use its given mass and divide it by the molar mass of ammonia (NH₃), which is 17.03 g/mol. This gives us the actual number of NH₃ molecules in the solution.
  • In this exercise step, 39.16 grams of NH₃ translates to 2.30 moles.
  • These moles are fundamental in calculating the molarity of the solution, as molarity is defined as the moles of solute per liter of solution.
By understanding moles of solute, you can confidently convert between mass and volume metrics, enabling precise concentration calculations critical in laboratory and industrial contexts.

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Most popular questions from this chapter

Mercury(II) nitrate is treated with hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), forming a precipitate and a solution. Write the molecular equation and the net ionic equation for the reaction. An acid is formed; is it strong or weak? Name each of the products. If \(81.15 \mathrm{~g}\) of mercury(II) nitrate and \(8.52 \mathrm{~g}\) of hydrogen sulfide are mixed in \(550.0 \mathrm{~g}\) of water to form \(58.16 \mathrm{~g}\) of precipitate, what is the mass of the solution after the reaction?

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