Chapter 12: Problem 60
In order to conduct three experiments that required different amounts of chloride ions, what mass of a \(0.150 \mathrm{~m}\) \(\mathrm{NaCl}\) solution is needed to obtain (a) \(0.00100 \mathrm{~mol} \mathrm{Cl}^{-}\) (b) \(0.00500 \mathrm{~mol} \mathrm{Cl}\), (c) \(0.0200 \mathrm{~mol} \mathrm{Cl}^{-}\) ?
Short Answer
Step by step solution
Understanding the Molarity Concept
Calculating the Required Volume for 0.00100 mol Cl-
Calculating Mass for 0.00100 mol Cl-
Repeat Steps 2 and 3 for 0.00500 mol Cl- and 0.0200 mol Cl-
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
\( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \).
Think of it like making lemonade; if you add more lemon juice (the solute) to the same amount of water, the lemonade becomes more concentrated or molar. To grasp this concept further, it's important to acknowledge that molarity calculations depend on the solute’s molar mass, the volume of the solution, and the desired concentration. Understanding molarity enables students to determine how much of a chemical to use to achieve a particular concentration, which is essential in performing accurate scientific experiments.
Concentration of Solutions
Moles of Solute
For molarity calculations, knowing the moles of solute present is foundational. This is because molarity describes the moles of solute per liter of solution. To calculate moles, one can use the molar mass of the substance (found on the periodic table) and the mass of the solute used, reinforcing the interconnected nature of these concepts in stoichiometry and solution preparation.
Stoichiometry
Stoichiometry also impacts molarity calculations. When you know the stoichiometry of a reaction, you can determine the amount of reactants needed to produce a desired amount of product, and vice versa. This allows for precise formulation of solutions. For instance, in the given exercise, the stoichiometry of sodium chloride (\(\mathrm{NaCl}\)) shows that one mole of \(\mathrm{NaCl}\) dissociates into one mole of sodium ions (\(\mathrm{Na}^{+}\)) and one mole of chloride ions (\(\mathrm{Cl}^{-}\)), enabling accurate calculations of the mass of \(\mathrm{NaCl}\) needed to get a required number of moles of chloride ions for various experiments.