By dissolving \(10 \mathrm{~g}\) of a non-volatile solute in \(100 \mathrm{~g}\) of benzene, the boiling point rises by \(1^{\circ} \mathrm{C}\). The molecular mass of solute is \(\left[\mathrm{K}_{\mathrm{b}}\right.\) for benzene \(\left.=2.53 \mathrm{~K} \mathrm{~m}^{-1}\right]\) (a) \(235 \mathrm{~g}\) (b) \(352 \mathrm{~g}\) (c) \(250 \mathrm{~g}\) (d) \(253 \mathrm{~g}\)

When you come across problems involving boiling point elevation in chemistry, understanding molality is key. Molality, often denoted by the letter 'm', measures the concentration of a solute in a solution. It's defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality isn't affected by temperature changes because it’s based on mass, not volume.

To calculate molality, you can use the formula: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]. This formula is especially useful when you're dealing with the colligative properties of solutions such as boiling point elevation or freezing point depression.

In the problem provided, we first need to determine the molality of the solution. By rearranging the boiling point elevation formula \[ \Delta T_b = K_b \times m \], we find \[ m = \frac{\Delta T_b}{K_b} \]. Thus, the calculation of molality becomes an essential step towards determining the molecular mass of the solute in subsequent steps.

The ebullioscopic constant (\(K_b\)), is a crucial factor when examining the phenomenon of boiling point elevation. Each solvent has its unique ebullioscopic constant value that represents the boiling point elevation per molal for a given solvent. It's often expressed in units of degrees Celsius per molal (\( ^\circ C\cdot m^{-1} \)).

Why is it important? This constant enables us to predict the change in boiling point when a non-volatile solute is dissolved in the solvent. In the exercise, the ebullioscopic constant for benzene is given as \(2.53\, K\cdot m^{-1}\). With this value, we can directly relate the change in boiling point to the molality of the solution, revealing the relationship between a solution’s concentration and its physical properties.

Determining the molecular mass of a solute is a fundamental skill in the field of chemistry, particularly when studying solutions. Once you have the molality and the mass of the solute, you can find the molecular mass. The molecular mass is the mass of one mole of a substance, expressed in grams per mole (\( g/mol \)).

From the molality, you can calculate the number of moles by using the formula \[ \text{moles} = m\times\frac{\text{mass of solvent (kg)}}{\text{molality}} \]. Then, the molecular mass is found by dividing the mass of the solute by the calculated number of moles \( \text{Molecular mass} = \frac{\text{mass of the solute (g)}}{\text{moles of the solute}} \). In the provided problem, this process enables the determination of the unknown molecular mass of the solute using the collected data and the known ebullioscopic constant.

Colligative properties are properties of solutions that depend on the ratio of solute particles to solvent molecules, regardless of the solute’s identity. Boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure are all colligative properties.

These properties are influenced by the number of particles in a solution which is directly correlated with its concentration. For boiling point elevation and freezing point depression, we use the molality of the solution along with constants specific to the solvent (\(K_b\) or \(K_f\)) to calculate how much these points change.

Understanding colligative properties is essential because they apply to many real-world scenarios, such as antifreeze in a car's engine or salting roads in winter to prevent ice formation. The exercise we are examining is a direct application of colligative properties, helping to illustrate how solutions behave differently from pure solvents.