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Commercial aqueous nitric acid has a density of \(1.42 \mathrm{~g} / \mathrm{mL}\) and is \(16 \mathrm{M}\). Calculate the percent \(\mathrm{HNO}_{3}\) by mass in the solution.

Short Answer

Expert verified
The percent of \(\mathrm{HNO}_{3}\) by mass in the solution is approximately \(71.0 \%\).

Step by step solution

01

Find the mass of \(\mathrm{HNO}_{3}\) in \(1 \mathrm{L}\) of solution

We have the molarity, which is the moles of solute per liter of solution, so let's find the mass of \(\mathrm{HNO}_{3}\) in \(1 \mathrm{L}\) of the solution. The molar mass of \(\mathrm{HNO}_{3}\) is \(1.01 + 14 + 3(16.00) = 63.01 \mathrm{g/mol}\) Mass of \(\mathrm{HNO}_{3}\) = (Molarity of \(\mathrm{HNO}_{3}\)) × (Molar mass of \(\mathrm{HNO}_{3}\)) × (Volume of solution) Mass of \(\mathrm{HNO}_{3}\) = \(16 \mathrm{M} \times 63.01 \mathrm{g/mol} \times 1 \mathrm{L}\) = \(1008.16 \mathrm{g}\)
02

Find the mass of \(1 \mathrm{L}\) of solution

We have the density of the solution, so we can calculate the mass of \(1 \mathrm{L}\) of solution. Mass of solution = Density × Volume Mass of solution = \(1.42 \mathrm{g/mL} \times 1000 \mathrm{mL}\) = \(1420 \mathrm{g}\)
03

Calculate the percent mass of \(\mathrm{HNO}_{3}\)

Now that we have the mass of \(\mathrm{HNO}_{3}\) and the total mass of the solution, we can calculate the percentage by mass. Percent mass of \(\mathrm{HNO}_{3}\) = \(\frac{Mass \: of \: \mathrm{HNO}_{3}}{Total\: mass \: of\: solution} \times 100\) Percent mass of \(\mathrm{HNO}_{3}\) = \(\frac{1008.16 \mathrm{g}}{1420 \mathrm{g}} \times 100 \approx 71.0 \%\) Hence, the percent of \(\mathrm{HNO}_{3}\) by mass in the solution is approximately \(71.0 \%\).

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Key Concepts

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

Density and Its Role in Solutions
Density is a fundamental concept in chemistry that describes how much mass is contained within a specific volume of a substance. It is typically expressed in grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³). In our exercise, density was used to calculate the mass of a given volume of solution. This is crucial for determining the overall composition of a chemical solution.
To calculate density, you can use the formula:
  • Density = Mass / Volume
Knowing the density of a substance allows chemists to understand how particles are packed within a given space. It also helps in converting between volume and mass, which is particularly useful when preparing chemical solutions.
Understanding Molar Mass
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It serves as a bridge between the microscopic world of atoms and molecules and the macroscopic measurements we can perform in the lab. For nitric acid ( HNO₃), its molar mass is the sum of the atomic masses of its constituent elements.
To calculate the molar mass of HNO₃, you add the atomic masses of hydrogen, nitrogen, and oxygen:
  • Hydrogen (H): 1.01 g/mol
  • Nitrogen (N): 14.00 g/mol
  • Oxygen (O): 16.00 g/mol x 3 = 48.00 g/mol
Thus, the molar mass of HNO₃ is 63.01 g/mol. Molar mass is a crucial factor in converting between moles and grams, allowing you to work with quantities in chemical reactions.
Molarity in Solution Chemistry
Molarity is a way of expressing concentration, specifically the number of moles of solute per liter of solution. It is denoted by M and is essential for accurately preparing and analyzing chemical solutions. For example, a 16 M nitric acid solution means there are 16 moles of HNO₃ in every liter of solution.
Understanding molarity helps with:
  • Calculating how much solute is needed to achieve a desired concentration.
  • Determining the amount of chemical that will react in reactions.
Essentially, molarity connects the volume of the solution with the number of molecules, bridging the gap between physical quantities and their chemical implications.
Solution Chemistry and Percent by Mass
Solution chemistry is the study of how substances dissolve and interact in a mixture. One key aspect is the concentration of components within a solution. Percent by mass refers to the mass of a solute divided by the total mass of the solution, presented as a percentage. It provides a straightforward metric for expressing how much of a solution is composed of a particular substance.
In the exercise, to find the percent by mass of HNO₃, we divided the mass of HNO₃ by the total mass of the solution (using the density to find this total mass), yielding about 71%. This measure is essential in many fields, from pharmacology to engineering, where precise formulations are necessary.
By understanding percent by mass, one can ensure correct dosage and reaction rates, making it an invaluable concept in solution chemistry.

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Most popular questions from this chapter

(a) Why is there no colloid in which both the dispersed substance and the dispersing substance are gases? (b) Michael Faraday first prepared ruby-red colloids of gold particles in water that were stable for indefinite times. \(\infty 00\) (Section 12.6) To the unaided eye these brightly colored colloids are not distinguishable from solutions. How could you determine whether a given colored preparation is a solution or colloid?

(a) What is an ideal solution? (b) The vapor pressure of pure water at \(60^{\circ} \mathrm{C}\) is 149 torr. The vapor pressure of water over a solution at \(60^{\circ} \mathrm{C}\) containing equal numbers of moles of water and ethylene glycol (a nonvolatile solute) is 67 torr. Is the solution ideal according to Raoult's law? Explain.

A "canned heat" product used to warm chafing dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin that has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{54}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin in formulating the mixture if the vapor pressure of ethanol at \(35^{\circ} \mathrm{C}\) over the mixture is to be 8 torr? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the solubility of \(\mathrm{N}_{2}\) in water in contact with air at ordinary atmospheric pressure \((1.0 \mathrm{~atm})\) is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, which is essentially an aqueous solution. At a depth of \(100 \mathrm{ft}\) in water, the pressure is \(4.0 \mathrm{~atm}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

A car owner who knows no chemistry has to put antifreeze in his car's radiator. The instructions recommend a mixture of \(30 \%\) ethylene glycol and \(70 \%\) water. Thinking he will improve his protection he uses pure ethylene glycol. He is saddened to find that the solution does not provide as much protection as he hoped. Why not?

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