A flask containing \(825 \mathrm{~mL}\) of a solution containing \(0.355 \mathrm{~mol}\) of \(\mathrm{CuSO}_{4}\) was left open overnight. The next morning the flask only contained \(755 \mathrm{~mL}\) of solution. What is the concentration (molarity) of the \(\mathrm{CuSO}_{4}\) solution remaining in the flask?

Understanding the concentration of a solution is crucial for numerous chemical applications, from laboratory experiments to industrial processes. It represents how much of a specific solute is present in a given amount of solvent. When we discuss the concentration, it usually refers to molarity, which is the number of moles of a substance per liter of solution.

In the context of the textbook exercise, molarity becomes an especially important concept when assessing changes in solution volume due to evaporation. Molarity is calculated with the formula: \[ M = \frac{n}{V} \]where \( M \) is the molarity, \( n \) is the number of moles, and \( V \) is the volume in liters. To make sense of molarity, consider it as a measure of 'solute density' within a solution—the greater the molarity, the more 'dense' the solution is with the solute.

At the heart of the molarity calculation lie the concepts of moles and volume. A mole is a fundamental unit in chemistry, denoting the amount of substance containing as many particles (atoms, molecules, ions, or electrons) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro’s number, approximately \(6.022 \times 10^{23}\). Conceptually, it's akin to a 'chemical dozen,' providing a convenient way to count particles at the atomic scale.

The volume of a solution is typically measured in liters (L) in the context of molarity calculations. When we measure the moles of solute dissolved in a specific volume of solvent, we're effectively examining the ratio of solute particles to the space they occupy. The precise relationship between moles and volume anchors our understanding of solution concentration.

Evaporation is a process where liquid turns into vapor, which can lead to a decrease in the volume of a chemical solution. However, it's important to note that only the solvent, typically water, evaporates, while the solute, such as \(\mathrm{CuSO}_4\), remains behind. As our textbook exercise illustrates, the solute's moles stay unchanged, but the volume of the solvent decreases due to evaporation.

This change in volume has a direct effect on concentration - as volume decreases, molarity increases. This is because molarity is inversely proportional to volume when the number of moles stays the same. It's a bit like a crowded elevator; as people (volume) leave, the remaining individuals (moles of solute) have less space, thus the elevator (solution) becomes more concentrated. The exercise demonstrates how evaporative processes can be analyzed to determine changes in solution concentration, without altering the absolute amount of solute.