A solution of \(5.36 \mathrm{~g}\) of a molecular compound dissolved in \(76.8 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) has a boiling point of \(81.48^{\circ} \mathrm{C}\). What is the molar mass of the compound? Boiling point for benzene \(=80.1^{\circ} \mathrm{C}\) \(K_{b}\) for benzene \(=2.53^{\circ} \mathrm{C} \mathrm{kg}\) solvent \(/ \mathrm{mol}\) solute

Boiling point elevation is a fascinating concept in solution chemistry. It describes the phenomenon where the boiling point of a solvent increases when a solute is dissolved in it.
This happens because adding a solute to a solvent decreases the solvent's vapor pressure, requiring a higher temperature to reach the boiling point.
We calculate boiling point elevation (\text{Δ}T_b) through the equation: \[ \triangle T_b = K_b \times \text{molality} \] In this exercise, we found \text{Δ}T_b by subtracting the normal boiling point of benzene (80.1\text{°}C) from the measured boiling point (81.48\text{°}C), resulting in \text{Δ}T_b = 1.38\text{°}C.
This step is crucial to solve problems involving the colligative properties of solutions, such as boiling point elevation.

Molality (\text{m}) is a measure of the concentration of a solute in a solution. It is defined as moles of solute per kilogram of solvent.
Unlike molarity, which is based on the volume of solvent, molality depends solely on the mass, making it useful for colligative property calculations where temperature changes are involved.
The formula for molality is:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{kg of solvent}} \] In the given exercise, we rearranged the boiling point elevation formula to solve for molality:
\[ \text{molality} = \frac{\text{Δ}T_b}{K_b} = \frac{1.38\text{°}C}{2.53\text{°}C \text{ kg/mol}} = 0.545 \text{ mol/kg} \] This value indicates the amount of solute causing the elevation in boiling point.

The molar mass of a compound is a fundamental property that represents the mass of one mole of its molecules. It is crucial in chemical calculations, particularly for stoichiometry and solution chemistry.
To determine molar mass from colligative properties, we use the formula:
\[ \text{Molar Mass} = \frac{\text{Mass of solute}}{\text{Moles of solute}} \] By calculating the molality and determining the moles of solute, we found the molar mass in the example problem. First, we found the moles of solute using:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
\[ \text{moles of solute} = \text{molality} \times \text{kg of solvent} = 0.545 \text{ mol/kg} \times 0.0768 \text{ kg} = 0.0418 \text{ mol} \]
Finally, by substituting the mass of the solute and moles, we get:
\[ \text{Molar Mass} = \frac{5.36 \text{ g}}{0.0418 \text{ mol}} = 128.23 \text{ g/mol} \] This value tells us the mass of one mole of the molecular compound.

Solution chemistry is a branch of chemistry focused on studying solutions, which are homogeneous mixtures composed of two or more substances. The main components include a solute and a solvent.
Some important concepts in solution chemistry are:

• Solutions and Concentration: Understanding solute-solvent interactions and how concentration defines the amount of solute in a given amount of solvent.
• Colligative Properties: Properties that depend on the number of particles in a solution, not their nature, such as boiling point elevation and freezing point depression.
• Solution Preparation: Techniques to create solutions with precise concentrations, including dissolving solutes and mixing.

In the exercise, we dissolved a molecular compound in benzene to study the boiling point elevation. This demonstrates principles of solution chemistry applied to real-world problems.
By understanding these core concepts, you'll gain a deeper insight into the behavior of solutions and their properties, making it easier to tackle various chemical problems.