A 325-mL sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\). (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

Molarity, often denoted by the symbol M, is a unit of concentration in chemistry that measures the number of moles of a solute per litre of solution. This fundamental concept is central to understanding solutions and their chemical reactions. The formula to calculate molarity is:
\[ M = \frac{\text{moles of solute}}{\text{Liters of solution}} \]
Imagine making lemonade; you care about how much lemon juice (the solute) goes into a certain volume of water (the solution). Similarly, in chemistry, if we dissolve a substance like calcium chloride \( \mathrm{CaCl_2} \), we need to know how much of it is in a certain volume of water to understand the solution's strength.
For instance, a solution with a molarity of 1M means that there is one mole of solute in one liter of solution. Higher molarity corresponds to a more concentrated solution. In our textbook exercise, calculating the molar concentration of chloride ions \( \mathrm{Cl^-} \) helps students understand how concentration changes with the amount of solute and volume of the solution.

Moles are the bridge between the microscopic world of atoms and the macroscopic world we live in. A mole is simply a unit that signifies a very large number of particles, specifically Avogadro's number \(6.02 \times 10^{23}\) of particles. In chemistry, counting atoms directly is practically impossible, so moles provide a convenient way to translate between atoms and bulk quantities.
\[ 1 \text{ mole} = 6.02 \times 10^{23} \text{ particles} \]
When you weigh out 35.45 grams of chlorine gas \( \mathrm{Cl_2} \), you actually have one mole, or \(6.02 \times 10^{23}\) molecules of chlorine. So, if a compound like calcium chloride \( \mathrm{CaCl_2} \) is used, we calculate the moles of each element in the compound to understand the quantities involved in the reaction. The step by step solution from the exercise illustrates how to convert the mass of a compound into moles, which is a fundamental skill in stoichiometry and a cornerstone in chemistry problem solving.

Stoichiometry is like the recipe for a chemical reaction. It tells you how much of each reactant you need and how much of each product you'll get. This process involves using the coefficients from the balanced chemical equation to convert between moles of different substances. Essentially, stoichiometry is the quantitative aspect of chemical reactions.
In the context of the exercise, understanding stoichiometry allows students to grasp the mole-to-mole ratios between different compounds and ions involved. The exercise provided shows that for every mole of \( \mathrm{CaCl_2} \), there are two moles of \( \mathrm{Cl^-} \) ions. This 1:2 ratio is taken directly from the formula of the compound and is pivotal in calculating the molarity of the solution, as well as determining how much of each ion is present in different volumes of the solution.

Solving chemistry problems often includes a set of systematic steps. First, you need to understand the problem: what information is given and what's asked for. Next, devise a plan using concepts like moles, molarity, and stoichiometry. This could involve setting up equations or converting units. Thirdly, execute the plan; perform the calculations. Lastly, review the results; ensure that they make sense in the context of the question.
When approaching a problem such as determining the concentration of ions in a solution, it's vital to apply these problem-solving skills. By accurately interpreting the formula of a compound, using stoichiometry to establish relationships between reactants and products, and calculating molarity, students can systematically approach and solve complex chemistry problems. The exercise critique advises that explaining the problem-solving process helps students understand not just the 'how,' but also the 'why' behind each step, leading to better long-term comprehension.