Determine the amount of sucrose in each solution. (a) \(48 \mathrm{~g}\) of a solution containing \(3.7 \%\) sucrose by mass (b) \(103 \mathrm{mg}\) of a solution containing \(10.2 \%\) sucrose by mass (c) \(3.2 \mathrm{~kg}\) of a solution containing \(14.3 \%\) sucrose by mass

When we talk about solution concentration, we're referring to the amount of a substance, called a solute, that is dissolved in a specific volume of solvent to form a solution. The solute in our examples is sucrose, a common sugar, and the solvent could be any liquid in which sucrose can dissolve, commonly water.

Concentration can be expressed in various ways, including mass percent, molarity, or molality. In our textbook exercise, the concentration of sucrose is given as a mass percent, which indicates the mass of sucrose in proportion to the total mass of the solution, including the sucrose and the solvent.

Mass percent is a very approachable way to express the concentration of a solution. It's calculated by dividing the mass of the solute by the total mass of the solution and then multiplying by 100 to get a percentage. For example, if a solution has a mass percent of 3.7%, this means that there are 3.7 grams of solute for every 100 grams of solution.

By using the mass percent, we can quickly calculate the mass of the solute in any given sample of the solution, which is particularly useful in food chemistry, pharmacology, and in diluting concentrated solutions to a desired concentration.

Stoichiometry is at the heart of the quantitative analysis in chemistry. It involves calculations based on the relationships between reactants and products in a chemical reaction, molar ratios from the balanced chemical equation, or, as in our exercise, percentages indicating composition of mixtures.

Understanding stoichiometry is crucial for precise manipulation and measurement of chemical substances in reactions. It allows chemists to predict how much product will result from a given amount of reactant or how much reactant is needed to make a given amount of product. Although stoichiometry often involves moles and molar relationships, it also applies to direct mass calculations like those in our exercise.

Problem solving in chemistry combines understanding of theoretical concepts and the practical application of mathematical calculations. To effectively solve problems like the ones in our textbook exercise, a student must carefully read the question, identify the known variables, select the appropriate formula, execute accurate conversions (such as turning percentages into decimals), perform the correct mathematical operations, and report the result using the proper unit of measurement.

Clarity and precision are vital; a small mistake in calculation can lead to a significantly wrong answer. Developing strong problem-solving skills in chemistry not only helps with academic exercises but is essential for real-world applications such as drug formulation, environmental monitoring, and material design.