Open Question
You can purchase nitric acid in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g>mL. Describe exactly how you would prepare 1.15 L of 0.100 M HNO3 from the concentrated solution.
Ch.13 - Solutions
Chapter 13, Problem 62c
Describe how to prepare each solution from the dry solute and the solvent. c. 125 g of 1.0% NaNO3 solution by mass
Verified step by step guidance1
Step 1: Understand the problem. We are asked to prepare a 1.0% by mass NaNO3 solution. This means that in every 100 g of the solution, there is 1 g of NaNO3.
Step 2: Calculate the total mass of the solution. We want to prepare 125 g of this solution. Since the solution is 1.0% NaNO3 by mass, this means that 1.0% of the total mass of the solution is NaNO3.
Step 3: Calculate the mass of NaNO3 needed. To find the mass of NaNO3 needed, multiply the total mass of the solution by the percentage of NaNO3 (expressed as a decimal).
Step 4: Measure out the calculated mass of NaNO3. Use a balance to measure out the correct mass of NaNO3.
Step 5: Dissolve the NaNO3 in the solvent. Add the measured NaNO3 to the solvent (usually water) and stir until it is completely dissolved. The total mass of the solution (NaNO3 + solvent) should be 125 g.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Mass Percent Concentration
Mass percent concentration is a way to express the concentration of a solution, defined as the mass of solute divided by the total mass of the solution, multiplied by 100. In this case, a 1.0% NaNO3 solution means that there is 1.0 g of NaNO3 in every 100 g of the solution. Understanding this concept is crucial for calculating the amount of solute needed to achieve the desired concentration.
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Mass Percent Calculation
Solution Preparation
Preparing a solution involves dissolving a specific amount of solute in a solvent to achieve a desired concentration. For a 1.0% NaNO3 solution, you would need to calculate the mass of NaNO3 required for the total mass of the solution (125 g in this case) and then add enough solvent (usually water) to reach that total mass. This process ensures that the solution has the correct concentration.
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Dilution and Mixing
Dilution is the process of reducing the concentration of a solute in a solution, typically by adding more solvent. When preparing a solution, it is important to mix the solute thoroughly with the solvent to ensure uniform distribution. Proper mixing is essential for achieving the desired concentration throughout the entire solution, which is particularly important in quantitative chemical applications.
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Related Practice
Open Question
Describe how to prepare each solution from the dry solute and the solvent. a. 1.00×102 mL of 0.500 M KCl b. 1.00×102 g of 0.500 m KCl c. 1.00×102 g of 5.0% KCl solution by mass
Textbook Question
Describe how to prepare each solution from the dry solute and the solvent. b. 125 g of 0.100 m NaNO3
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Open Question
For a solution prepared by dissolving 28.4 g of glucose (C6H12O6) in 355 g of water with a final volume of 378 mL, calculate the concentration in each unit: a. molarity, b. molality, c. percent by mass, d. mole fraction, e. mole percent.
Textbook Question
A solution is prepared by dissolving 20.2 mL of methanol (CH3OH) in 100.0 mL of water at 25 °C. The final volume of the solution is 118 mL. The densities of methanol and water at this temperature are 0.782 g/mL and 1.00 g/mL, respectively. For this solution, calculate the concentration in each unit. a. molarity
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Textbook Question
A solution is prepared by dissolving 20.2 mL of methanol (CH3OH) in 100.0 mL of water at 25 °C. The final volume of the solution is 118 mL. The densities of methanol and water at this temperature are 0.782 g/mL and 1.00 g/mL, respectively. For this solution, calculate the concentration in each unit. b. molality
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