Problem 105
Question
Concentrated aqueous ammonia is \(14.8 \mathrm{M}\) and has a density of \(0.90 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the weight percent of \(\mathrm{NH}_{3}\) in the solution.
Step-by-Step Solution
Verified Answer
The weight percent of \(\mathrm{NH}_{3}\) is approximately 27.96%.
1Step 1: Calculate the Volume of One Liter of Solution
To find the mass of the solution per liter, we use the density. Since the density is given as \(0.90 \text{ g/cm}^3\), this is equivalent to \(0.90 \text{ g/mL}\). Therefore, the mass of 1 L (1000 mL) of solution is \(0.90 \times 1000 = 900 \text{ g}\).
2Step 2: Calculate Moles of NH3 in Solution
The molarity (M) indicates moles of solute per liter of solution. Given the solution is \(14.8 \text{ M}\), this means there are 14.8 moles of \(\text{NH}_3\) in 1 liter (1000 mL) of solution.
3Step 3: Calculate Mass of NH3
To find the mass of \(\text{NH}_3\), multiply the number of moles by the molar mass of \(\text{NH}_3\). The molar mass of \(\text{NH}_3\) is \(14 + 3 \times 1 = 17 \text{ g/mol}\). Thus, the mass of \(\text{NH}_3\) is \(14.8 \times 17 = 251.6 \text{ g}\).
4Step 4: Calculate Weight Percent of NH3
Weight percent is calculated as \((\text{mass of solute} / \text{mass of solution}) \times 100\). The mass of \(\text{NH}_3\) is 251.6 g, and the mass of the solution is 900 g. The weight percent is thus \((251.6 / 900) \times 100 \approx 27.96\%\).
Key Concepts
MolarityDensityMolar MassAqueous Ammonia Solution
Molarity
Molarity is a way of measuring the concentration of a solution. It's defined as the number of moles of solute (the substance being dissolved) per liter of solution. This measurement helps chemists understand how much of a substance is present in a given volume of liquid.
To calculate molarity, use the formula:
To calculate molarity, use the formula:
- Molarity (M) = Moles of solute / Liters of solution
Density
Density tells us how much mass is in a specific volume of a substance. It's a crucial concept in chemistry and physics because it helps convert between mass and volume, especially when dealing with solutions.
The formula to calculate density is:
The formula to calculate density is:
- Density = Mass / Volume
Molar Mass
Molar mass is the mass of one mole of a given substance. It is usually expressed in grams per mole (g/mol). The molar mass allows you to convert between the mass of a substance and the number of moles, which is essential for stoichiometric calculations and understanding chemical reactions.
To find the molar mass of a compound, add up the atomic masses of all atoms within it.
For ammonia (\(\text{NH}_3\)), the molar mass is calculated as:
To find the molar mass of a compound, add up the atomic masses of all atoms within it.
For ammonia (\(\text{NH}_3\)), the molar mass is calculated as:
- Nitrogen: 14 g/mol
- Hydrogen: 3 x 1 g/mol = 3 g/mol
- Total molar mass of NH₃ = 14 + 3 = 17 g/mol
Aqueous Ammonia Solution
Aqueous ammonia solution is ammonia dissolved in water. It is commonly used in various industrial and household cleaning products. Understanding its properties, such as molarity and density, is important for its application and safe handling.
In this exercise, we're dealing with a concentrated aqueous ammonia solution with a known molarity of 14.8 M. This indicates a very high concentration of ammonia in the water, making it potent for cleaning and industrial use.
When calculating its properties, such as weight percent, you consider both the ammonia and the water, along with their combined mass and volume. This enables the safe and effective utilization of ammonia solution in different products.
In this exercise, we're dealing with a concentrated aqueous ammonia solution with a known molarity of 14.8 M. This indicates a very high concentration of ammonia in the water, making it potent for cleaning and industrial use.
When calculating its properties, such as weight percent, you consider both the ammonia and the water, along with their combined mass and volume. This enables the safe and effective utilization of ammonia solution in different products.
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