A mixture of \(\mathrm{NaCl}\) and sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) of combined mass \(10.2 \mathrm{~g}\) is dissolved in enough water to make up a \(250 \mathrm{~mL}\) solution. The osmotic pressure of the solution is 7.32 atm at \(23^{\circ} \mathrm{C}\). Calculate the mass percent of \(\mathrm{NaCl}\) in the mixture.

Colligative properties are characteristics of solutions that depend on the number of particles in a given volume of solvent and not on the type of particles. Osmotic pressure is one such colligative property which is particularly useful when determining the concentration of a solution. As seen in the textbook exercise, osmotic pressure (\( \text{Π} \)) can be used to calculate the moles of solute in a solution using the formula \( \text{Π} = \frac{nRT}{V} \), where \( n \) represents the number of moles of solute, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the solution in liters.

In the context of the exercise, since osmotic pressure is influenced by the total number of particles, both \( \mathrm{NaCl} \) and sucrose contribute equally to the total pressure if they dissociate into the same number of particles. This property is exploited to solve for the individual components of a multi-solute solution.

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is a fundamental concept in chemistry that can be used to relate the mass of a substance to the number of moles it contains, as indicated by the formula \( \text{mass} = \text{moles} \times \text{molar mass} \). In the osmotic pressure exercise, knowledge of the molar mass of \( \mathrm{NaCl} \) enables the calculation of its mass in the solution after determining the number of moles. By definition, the molar mass of a compound is the sum of the masses of its constituent elements, derived from the periodic table, which reflects the compound’s composition.

The mass percent composition is a way of expressing the concentration of an individual component of a mixture or solution. It is calculated using the mass of the component of interest divided by the total mass of the mixture, multiplied by 100%. In formulas, mass percent (\% ) = \( \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100 \% \).

For the solution mentioned in the exercise, mass percent composition allows us to determine the proportion of \( \mathrm{NaCl} \) once its mass is known. This proportion is the mass of \( \mathrm{NaCl} \) divided by the total mass of the mixture (10.2g), and then multiplied by 100 to give a percentage. This figure is useful when considering how the components in a mixture contribute to its overall properties.

Solutions and mixtures are different forms of combinations of substances. A solution is a homogeneous mixture where one substance (the solute) is dissolved in another (the solvent) at the molecular or ionic level, resulting in a single phase. The exercise illustrates the process of dissolving both \( \mathrm{NaCl} \) and sucrose into water to create a solution.

One of the remarkable phenomena of solutions is their ability to have different components yet present as a uniform single-phase system. This homogeneity allows for the colligative properties, such as osmotic pressure, to be applicable in calculations regarding the solution's composition. Understanding how solutes interact within a solvent is key to grasping the complexity of how mixtures behave differently than their individual components.