A \(0.500-\mathrm{g}\) sample of a compound is dissolved in enough water to form \(100.0 \mathrm{~mL}\) of solution. This solution has an osmotic pressure of \(2.50\) atm at \(25^{\circ} \mathrm{C}\). If each molecule of the solute disso- ciates into two particles (in this solvent), what is the molar mass of this solute?

Understanding the molar mass of a substance is essential in chemistry, especially when dealing with solutions. The molar mass is the weight in grams of one mole of a substance, which is numerically equivalent to its molecular weight in atomic mass units (amu). To calculate it, you sum the atomic weights of all atoms in the molecule. For instance, water (H₂O) has a molar mass of approximately 18.02 g/mol, since each of the two hydrogen atoms contributes 1.01 g/mol, and the oxygen atom contributes around 16.00 g/mol.

In our exercise, we find the molar mass by dividing the given mass of the dissolved compound by the number of moles of solute. This value will assist us in identifying the compound or at least verifying its purity. Remember, the accuracy of your molar mass calculation can significantly impact the results of any further calculations that use this value, such as determination of solution concentration and calculation of osmotic pressure.

Colligative properties are physical properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. They are essential for understanding how solutes affect the physical properties of solvents.

In the problem at hand, osmotic pressure is the colligative property we focus on. Osmotic pressure is the pressure that must be applied to a solution to prevent the inward flow of water across a semipermeable membrane. It's determined by the concentration of solute particles and temperature, which is why a correct calculation of the molar concentration directly influences the computation of osmotic pressure.

The van't Hoff factor (i) is a dimensionless quantity used in calculations involving colligative properties. It describes the number of particles that a compound dissociates into when it dissolves in a solvent. For example, a van't Hoff factor of 1 would signify a non-dissociating molecule like glucose, whereas a factor of 2 indicates a compound like NaCl, which dissociates into two particles: a sodium ion (Na⁺) and a chloride ion (Cl^–).

In our textbook exercise, the solute dissociates into two particles, so the van't Hoff factor is 2. This information is crucial because it means that the observed osmotic pressure is due to twice the number of particles in the solution than if the solute didn't dissociate. Therefore, the presence of a van't Hoff factor affects both the molar concentration and the molar mass calculations for the solute.

The concentration of a solution tells us how much solute is present in a given volume of solvent, and it is a measure of how 'dense' a solution is with solute particles. Common units of concentration include molarity (moles per liter), molality (moles per kilogram of solvent), and percent composition by mass or volume.

In terms of our exercise, the concentration of the solution is crucial for calculating both the osmotic pressure and molar mass. By dividing the osmotic pressure by the gas constant and the temperature in kelvins, we get the molarity of particles in the solution. However, because each particle dissociates, the actual concentration of the solute prior to dissociation is half of this value. Understanding the concept of solution concentration is critical when interpreting or predicting the behavior of solutions in various scientific and industrial processes.