Question

If 10. \(\mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?

Short answer

Using 10 g of AgNO3, we can prepare 235.6 mL of 0.25 M AgNO3 solution.

Step-by-step solution

Determine the number of moles of AgNO3 in 10 grams

To calculate the number of moles, we can use the formula: Number of moles = mass / molar mass First, we need to find the molar mass of AgNO3. The molar mass is the sum of the atomic masses of all the elements in the compound. In AgNO3, we have: 1 Silver (Ag) atom: 1 * 107.87 g/mol = 107.87 g/mol 1 Nitrogen (N) atom: 1 * 14.01 g/mol = 14.01 g/mol 3 Oxygen (O) atoms: 3 * 16.00 g/mol = 48.00 g/mol The molar mass of AgNO3 = 107.87 + 14.01 + 48.00 = 169.88 g/mol. Now, we can find the number of moles of AgNO3 in 10 g: Number of moles = 10 g / 169.88 g/mol = 0.0589 mol

Use the molarity formula to find the volume of the solution

We can find the volume of the solution using the molarity formula: Molarity (M) = number of moles / volume (L) We are given the molarity (0.25 M) and the number of moles (0.0589 mol). We can rearrange the formula to solve for the volume (L): Volume (L) = number of moles / molarity (M) Volume (L) = 0.0589 mol / 0.25 M = 0.2356 L To convert this to milliliters (mL), we can multiply by 1000: Volume (mL) = 0.2356 L * 1000 = 235.6 mL

Final Answer

Thus, 235.6 mL of 0.25 M AgNO3 solution can be prepared from 10 g of AgNO3.

Key concepts

Molar Mass

Understanding the concept of molar mass is fundamental in chemical calculations, as it bridges the gap between the microscopic world of atoms and molecules and the macroscopic world that we can measure. Molar mass, often expressed in grams per mole (g/mol), is the weight of one mole of a given substance. A mole, defined as Avogadro's number (\(6.022 \times 10^{23}\)), represents a set amount of particles, be it atoms, ions, or molecules.

To find the molar mass of a compound like silver nitrate (\(AgNO_3\)), you sum the atomic masses of all the atoms within a molecule. Here, we calculate it by adding the molar mass of one silver atom, one nitrogen atom, and three oxygen atoms. This sum represents the molar mass of one mole of silver nitrate, a crucial step in solving various chemical problems, including molarity calculations.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the law of conservation of mass and the concept of moles. Stoichiometry allows us to predict the amounts of substances consumed and produced in a given reaction. Techniques involving stoichiometric calculations include balancing chemical equations, determining limiting reactants, and calculating theoretical yields.

In the context of solution preparation, stoichiometry helps us connect the amount of solute (in moles) and the volume of solvent to achieve the desired molarity. Mastering stoichiometry is not only crucial for lab preparation but also for understanding the quantitative aspects of chemical reactions.

Solution Preparation

Preparing a chemical solution with a specific concentration requires careful measurement and calculation. Molarity (M), a common unit of concentration, is defined as the number of moles of solute per liter of solution. When preparing solutions, the steps typically involve dissolving a measured amount of solute (often in grams) into a solvent to achieve a desired volume. The process involves calculating the required mass of solute using molarity and volume information, followed by dissolving and diluting to the final volume.

For example, in laboratory settings, this might mean dissolving solid silver nitrate to create a solution of a certain molarity, ensuring precision for subsequent experimental procedures. Detailing the process step by step, as shown in the original exercise, is pivotal for learners to confidently prepare solutions independently.

Chemical Calculations

Chemical calculations encompass various mathematical techniques used in chemistry to deal with the quantities of substances involved in chemical reactions and solutions. This includes calculations related to molar mass, molarity, stoichiometry, percentage composition, empirical and molecular formulas, and more. These calculations are the backbone of experimental chemistry, allowing scientists and students to design experiments, analyze results, and predict outcomes.

To solve the problem at hand, it's essential to perform a series of calculations, starting with finding the molar mass of the compound, determining the moles of the compound, and finally calculating the volume of the solution using molarity. By understanding these chemical calculations, students can tackle a wide variety of problems in chemistry with the assurance that they have the quantitative aspect of their experiments under control.