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Chapter 4: Problem 47

How many grams of \(\mathrm{AgNO}_{3}\) are needed to prepare 300 \(\mathrm{mL}\) of a \(1.00 M\) solution?

Short Answer

Expert verified
You need 50.96 grams of \( \mathrm{AgNO}_3 \).

Step by step solution

01

Understand the Concept

To solve this problem, we need to understand the concept of molarity. Molarity, expressed as \( M \), is the number of moles of solute per liter of solution. We can use this relationship to find the moles of \( \mathrm{AgNO}_3 \).
02

Molarity Equation

Use the molarity equation: \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]. Since the molarity \( M \) is given as \( 1.00 \ M \) and the volume is \( 300 \ \mathrm{mL} \), we first convert the volume into liters, which is \( 0.300 \ \mathrm{L} \).
03

Calculate Moles of Silver Nitrate

Using the molarity equation, calculate the moles of \( \mathrm{AgNO}_3 \) needed: \[ 1.00 \ M = \frac{\text{moles of } \mathrm{AgNO}_3}{0.300} \] Solving for moles gives us: \[ \text{moles of } \mathrm{AgNO}_3 = 1.00 \times 0.300 = 0.300 \text{ moles} \]
04

Find Molar Mass of Silver Nitrate

Calculate the molar mass of \( \mathrm{AgNO}_3 \). The atomic masses are approximately \( \mathrm{Ag}: 107.87 \ \mathrm{g/mol} \), \( \mathrm{N}: 14.01 \ \mathrm{g/mol} \), \( \mathrm{O}: 16.00 \ \mathrm{g/mol} \). The molar mass is: \[ 107.87 + 14.01 + (16.00 \times 3) = 169.87 \ \mathrm{g/mol} \]
05

Calculate Grams of Silver Nitrate

Multiply the moles of \( \mathrm{AgNO}_3 \) by its molar mass to find the mass in grams: \[ 0.300 \text{ moles} \times 169.87 \ \mathrm{g/mol} = 50.96 \ \mathrm{g} \]
06

Conclusion

To prepare a 300 \( \mathrm{mL} \) solution of \( 1.00 \ M \) \( \mathrm{AgNO}_3 \), you need \( 50.96 \ \mathrm{grams} \) of \( \mathrm{AgNO}_3 \).

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Key Concepts

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

Molarity
Molarity is a way of expressing the concentration of a solution. It measures how much solute is present in a given volume of solution. This is crucial in chemistry because knowing the concentration allows chemists to predict how substances will react in solution. When we talk about molarity, we use the symbol \( M \), which is defined as the number of moles of solute per liter of solution. For example, a \( 1.00 \, M \) \( \mathrm{AgNO}_3 \) solution means there is 1 mole of \( \mathrm{AgNO}_3 \) in every liter of solution.
  • To calculate the molarity, use the formula: \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
  • Always convert volumes to liters when using this formula. This ensures consistency and accuracy in calculations.
Understanding molarity is fundamental to solution preparation, as it directly influences how much of a solute is needed to achieve a desired concentration.
Chemical Calculations
Chemical calculations are critical in determining how much of each substance is needed for reactions or solutions. These calculations ensure that reactions go to completion without excess reactants, an important consideration for both cost-efficiency and safety. For the exercise, chemical calculations help determine the mass of \( \mathrm{AgNO}_3 \) required.The problem involves calculating the moles of solute needed using the molarity formula and then converting moles to grams. Here鈥檚 how:
  • First, find the total moles needed using the molarity formula: \( M = \frac{\text{moles of \( \mathrm{AgNO}_3 \)}}{\text{liters of solution}} \).
  • Then multiply the moles by the molar mass of \( \mathrm{AgNO}_3 \) to convert to grams.
These calculations allow precise preparation of desired solution concentrations, a frequent task in chemistry labs.
Mole Concept
The mole concept is a fundamental principle in chemistry that helps quantify amounts of substances. It enables chemists to express amounts of a chemical substance in terms of the number of atoms, ions, or molecules rather than mass. One mole refers to \( 6.022 \times 10^{23} \) entities, known as Avogadro's number.This concept is especially useful in stoichiometry, where chemical reactions are balanced using moles. In the provided exercise, we need to determine how many moles of \( \mathrm{AgNO}_3 \) are needed to make a 1.00 \( M \) solution:
  • Start by converting milliliters to liters as molarity is defined per liter.
  • Use the mole concept to relate moles to grams via the molar mass, which acts as the bridge in these calculations.
Mastering this concept is crucial as it forms the basis for nearly all quantitative aspects of chemical reactions and solution preparation.

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Most popular questions from this chapter

Write the net ionic equation for the reaction, if any, that occurs on mixing (a) solutions of potassium bromide and silver nitrate. (b) a solution of nitric acid and calcium metal to produce calcium nitrate and hydrogen gas. (c) solutions of lithium hydroxide and iron(III) chloride.

A \(6.00-\mathrm{g}\) sample of sodium hydroxide is added to a 1.00-L volumetric flask, and water is added to dissolve the solid and fill the flask to the mark. A \(100-\mathrm{mL}\) portion of this solution is added to a 5.00-L volumetric flask, and water is added to fill the flask to the mark. What is the concentration of \(\mathrm{NaOH}\) in the second flask?

How many grams of barium chloride are needed to prepare \(1.00 \mathrm{~L}\) of a \(0.100 \mathrm{M}\) solution?

An aqueous sample is known to contain either \(\mathrm{Mg}^{2+}\) or \(\mathrm{Ba}^{2+}\) ions. Treatment of the sample with \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) produces a precipitate, but treatment with ammonium sulfate does not. Use the solubility rules (see Table 4.1) to determine which cation is present.

Although silver chloride is insoluble in water, adding ammonia to a mixture of water and silver chloride causes the silver ions to dissolve because of the formation of \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) ions. What is the concentration of \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) ions that results from the addition of excess ammonia to a mixture of water and \(0.022 \mathrm{~g}\) silver chloride if the final volume of the solution is \(150 \mathrm{~mL}\) ?
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