Question

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

Short answer

The volume of the 0.25M AgNO3 solution that can be prepared from 10g of AgNO3 is 0.2356 Liters.

Step-by-step solution

Calculate the molar mass of AgNO3

To calculate the molar mass of AgNO3, we add the molar masses of the elements that make up AgNO3: silver (Ag), nitrogen (N), and oxygen (O). Molar mass of AgNO3 = (molar mass of Ag) + (molar mass of N) + (3 × molar mass of O) The molar masses of each element are: Ag = 107.87 g/mol N = 14.01 g/mol O = 16.00 g/mol Molar mass of AgNO3 = (107.87 g/mol) + (14.01 g/mol) + (3 × 16.00 g/mol) = 169.87 g/mol

Convert mass of AgNO3 to moles

10 grams of AgNO3 is given. To find the moles of AgNO3, we can use the formula: moles = mass (g) / molar mass (g/mol) moles = 10 g / 169.87 g/mol = 0.0589 mol So, we have 0.0589 moles of AgNO3.

Use the molarity equation to find the volume of the solution

Now that we have the number of moles of AgNO3, we can use the molarity equation: Molarity (M) = moles of solute / volume of solution (in Liters) We want the solution's molarity to be 0.25M, and we have 0.0589 moles of AgNO3. Now we can solve for the volume: 0.25 M = 0.0589 mol / volume (L) volume (L) = 0.0589 mol / 0.25 M = 0.2356 L The volume of the 0.25M AgNO3 solution can be prepared from 10g of AgNO3 is 0.2356 Liters.

Key concepts

Molarity Calculation

Molarity is a way to express the concentration of a solution. It’s defined as the number of moles of solute (what you dissolve) per liter of solution. Understanding molarity helps you prepare chemical solutions accurately.

When calculating molarity, you use the formula:

• Molarity (M) = moles of solute / volume of solution in liters

In the exercise, we had 0.0589 moles of AgNO₃ and wanted a solution with a molarity of 0.25 M.
Using the molarity formula, we solved for the volume of the solution to find how much liquid we needed to achieve the desired concentration.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It serves as a bridge between the mass of a substance and the amount in moles.

To find the molar mass of a compound like AgNO₃, you sum the molar masses of its elements:

• Silver (Ag): 107.87 g/mol
• Nitrogen (N): 14.01 g/mol
• Oxygen (O): 16.00 g/mol

Adding these together gives:

• (107.87) + (14.01) + 3(16.00) = 169.87 g/mol

This value allows us to convert between grams and moles, which is crucial for preparing solutions.

Chemical Solutions

Chemical solutions are mixtures consisting of a solute dissolved in a solvent. They can vary in concentration, which describes how much solute is present per unit volume of solvent.

In the exercise, AgNO₃ is the solute, and it’s dissolved in a solvent to make a solution. Understanding how to calculate and create solutions of specific concentrations is a key skill in chemistry.

Accurate solution preparation ensures that reactions occur under optimal conditions, impacting everything from lab experiments to industrial processes.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the amounts of reactants and products involved in chemical reactions. It ensures that chemical reactions are understood and predicted with accuracy.

Although this exercise focuses on solution preparation, stoichiometry underpins the entire process. It helps determine the precise amounts of each reactant needed. This ensures reactions proceed efficiently, avoiding excess or shortage.

By knowing the molarity and volume, and through calculations similar to those in the exercise, stoichiometry keeps the reaction on track and optimizes resource use.