Nitric acid is purchased from chemical suppliers as a solution that is \(70 \% \mathrm{HNO}_{3}\) by mass. What mass (in grams) of a \(70 \% \mathrm{HNO}_{3}(\mathrm{aq})\) solution is required to prepare \(250 \mathrm{~g}\) of a \(2.0 \mathrm{~m} \mathrm{HNO}_{3}\) (aq) solution? The density of \(70 \% \mathrm{HNO}_{3}(\mathrm{aq})\) is \(1.42 \mathrm{~g} \cdot \mathrm{cm}^{-1}\).

Short Answer

Expert verified
To prepare 250 g of a 2.0 M HNO3 solution, 45.007 g (or 31.6929 cm3) of the 70% HNO3 solution is required.

Step by step solution

01

Understand Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. The molarity of the desired solution is given as 2.0 M, which means there are 2.0 moles of HNO3 in 1 liter of solution.
02

Calculate Moles of HNO3 Needed

To prepare 250 g of a 2.0 M solution, first find out how many moles of HNO3 are required. Since the density of water is approximately 1 g/cm3 and assuming the solution's density is close to water, 250 g of solution is about 250 mL or 0.250 L. Therefore we need 2.0 moles/L * 0.250 L = 0.5 moles of HNO3.
03

Calculate Mass of Pure HNO3 Needed

The molar mass of HNO3 is approximately 63.01 g/mol. Therefore, the mass of pure HNO3 needed is 0.5 moles * 63.01 g/mol = 31.505 g.
04

Calculate Mass of 70% HNO3 Solution Needed

Since the solution is 70% HNO3 by mass, this means that in every 100 g of solution, there are 70 g of pure HNO3. To find the mass of the 70% solution needed to get 31.505 g of pure HNO3, use the proportion: (70 g HNO3 / 100 g solution) = (31.505 g HNO3 / x g solution), solving for x, we get x = (31.505 g * 100 g) / 70 g = 45.007 g.
05

Convert grams to volume using density

To find out the volume of 70% solution that corresponds to the 45.007 g needed, use the density. The density is given as 1.42 g/cm3, which means each cm3 of the solution weighs 1.42 g. Therefore, the volume required is 45.007 g / 1.42 g/cm3 = 31.6929 cm3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). A mole is a standard unit of measurement in chemistry that represents a specific number of particles—Avogadro's number, approximately equal to \(6.022 \times 10^{23}\) atoms, ions, or molecules.

Imagine you have a puzzle, and each puzzle piece represents one molecule. The molar mass tells you how much one mole of these puzzle pieces would weigh. For example, the molar mass of nitric acid (\(HNO_3\)) is about 63.01 g/mol, which means if we gather Avogadro's number of \(HNO_3\) molecules together, they would weigh approximately 63.01 grams.

To calculate the molar mass, one adds the atomic masses of all the atoms in the molecule. The atomic masses can be found on the periodic table, and they must be added as per the number of times each atom appears in the molecule. For example, \(HNO_3\) contains one nitrogen (N), one hydrogen (H), and three oxygen (O) atoms, so its molar mass is calculated as follows:
Solutions Preparation

Calculating Ingredients for the Perfect Mixture

Preparing a chemical solution requires precise measuring of the solute (the substance to be dissolved) and the solvent (the liquid in which the solite is dissolved) to achieve a desired concentration. Think of it like following a recipe to bake a cake. You need the right proportion of ingredients mixed together to get a delicious result. Similarly, solutions preparation in chemistry follows a 'recipe' to achieve the desired concentration of the solution.

When preparing solutions, it is crucial to understand the solute's purity and the solvent's properties. For instance, if you have a solute with a high purity level, such as a concentrated nitric acid solution (70% \(HNO_3\)), to dilute it, you must calculate how much solute and solvent are needed to get to a specific molarity (a measure of concentration). The process might involve diluting a concentrated solution with water or mixing the solute directly with the solvent in a calibrated volumetric flask. With accurate measurements and proper mixing techniques, the desired concentration can be achieved.
Concentration of Solution
The concentration of a solution describes how much solute is present in a given quantity of solvent. Concentration can be expressed in several ways, including molarity, which is moles of solute per liter of solution (mol/L or M). Molarity reflects the strength of the solution.

Continuing with the cooking analogy, the molarity is akin to the intensity of a flavor in a dish. Just as you might adjust the level of seasoning to achieve the desired taste, in chemistry, you adjust the molarity to obtain the required solution potency for a reaction or experiment. For example, a 2.0 M solution of \(HNO_3\) means there are 2.0 moles of nitric acid for every liter of solution. If you need less 'flavor' or a less concentrated solution, you add more solvent. Conversely, if you need more 'flavor' or a higher concentration, you add more solute.

For students and scientists alike, understanding the relationship between the amount of solute, the volume of solvent, and the molarity is essential in concocting the right concentration for any particular application. By mastering the calculations associated with molarity, you can create a precise and effective environment for chemical reactions to occur.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A chemist studying the propertics of photographic emulsions needed to prepare \(25.00 \mathrm{~mL}\). of \(0.155 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) (aq). What mass of silver nitrate must be placed into a \(25.00-\mathrm{mL}\) volumctric flask and dissolved and diluted to the mark with water?

The sulfuric acid solution that is purchased for a stockroom has a molarity of \(17.8\) M; all sulfuric acid solutions for experiments are prepared by dilution of this stock solution. (a) Determine the volume of \(17.8 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) that must be diluted to \(250 \mathrm{~mL}\) to prepare a \(2.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) solution. (b) An experiment requires a \(0.50 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) solution. The stockroom manager estimates that \(6.0\) L of the acid is needed. What volume of \(17.8 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) must be used for the preparation?

A chemist prepared an aqueous solution by mixing \(2.50 \mathrm{~g}\) of ammonium phosphate trihydrate, \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) and \(1.50 \mathrm{~g}\) of potassium phosphate, \(\mathrm{K}_{3} \mathrm{PO}_{4}\), with \(500 \mathrm{~g}\) of water. (a) Determine the number of moles of formula units of each compound that was measured. (b) How many moles of \(\mathrm{PO}_{4}{ }^{3-}\) are present in solution? (c) Calculate the mass of phosphate ions present in the solution. (d) What is the total mass of the water present in the solution?

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

(a) What mass (in grams) of \(\mathrm{NaOH}\) must be mixed with \(250 \mathrm{~g}\) of water to prepare a \(0.22 \mathrm{~m} \mathrm{NaOH}\) solution? (b) Calculate the amount (in moles) of cthylene glycol, \(\mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{OH}\), that should be added to \(2.0 \mathrm{~kg}\) of water to prepare a \(0.44 \mathrm{~m} \mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{OH}\) (aq) solution. (c) Determine the amount (in moles) of HCl that must be dissolved in \(500 \mathrm{~g}\) water to prepare a \(0,0010 m\) HClaq) solution.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free