Problem 104 - Chemistry: The Central Science

Question

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) boils at \(46.30^{\circ} \mathrm{C}\) and has a density of \(1.261 \mathrm{~g} / \mathrm{mL}\). (a) When \(0.250 \mathrm{~mol}\) of a nondissociating solute is dissolved in \(400.0 \mathrm{~mL}\) of \(\mathrm{CS}_{2}\), the solution boils at \(47.46^{\circ} \mathrm{C}\). What is the molal boiling-point-elevation constant for \(\mathrm{CS}_{2}\) ? (b) When \(5.39 \mathrm{~g}\) of a nondissociating unknown is dissolved in \(50.0 \mathrm{~mL}\) of \(\mathrm{CS}_{2}\), the solution boils at \(47.08^{\circ} \mathrm{C}\). What is the molecular weight of the unknown?

Step-by-Step Solution

Answer

The molal boiling-point-elevation constant for CS₂ is 2.34 °C·kg/mol, and the molecular weight of the unknown solute is 112.4 g/mol.

Step 1: Calculate the mass of CS₂

First, we are given the volume of the solution (400.0 mL) and the density of CS₂ (1.261 g/mL). We can use these values to determine the mass of CS₂ in the solution: Mass of CS₂ = volume × density Mass of CS₂ = 400.0 mL × 1.261 g/mL = 504.4 g.

Step 2: Calculate the molality of the solution

Molality is defined as the moles of solute per kilogram of solvent. We are given the moles of the solute (0.250 mol) and have calculated the mass of CS₂ (504.4 g). We can convert the mass of CS₂ to kg and then calculate the molality of the solution: Mass of CS₂ in kg = 504.4 g × 1 kg/1000 g = 0.5044 kg Molality (m) = moles of solute / mass of solvent in kg Molality (m) = 0.250 mol / 0.5044 kg = 0.495 mol/kg. Step 2: Find the molal boiling-point-elevation constant

Step 3: Calculate the boiling point elevation

Boiling point elevation (∆Tb) is the difference between the boiling point of the pure solvent and the boiling point of the solution: ∆Tb = boiling point of solution - boiling point of pure solvent ∆Tb = 47.46 °C - 46.30 °C = 1.16 °C.

Step 4: Use boiling point elevation formula to find the molal boiling-point-elevation constant

The boiling point elevation formula is given as: ∆Tb = Kb × m Where, Kb = molal boiling-point-elevation constant m = molality of the solution ∆Tb = boiling point elevation We can rearrange the formula to find Kb: Kb = ∆Tb / m Kb = 1.16 °C / 0.495 mol/kg = 2.34 °C·kg/mol. Step 3: Find the molecular weight of the unknown solute

Step 5: Calculate the molality of the new solution

We are given that 5.39 g of the unknown solute is dissolved in 50.0 mL of CS₂. The boiling point of this new solution is 47.08 °C. First, we need to find the mass of CS₂ in this solution: Mass of CS₂ = volume × density Mass of CS₂ = 50.0 mL × 1.261 g/mL = 63.05 g Mass of CS₂ in kg = 63.05 g × 1 kg/1000 g = 0.06305 kg Since we don't yet know the molecular weight of the unknown solute, we can't find the moles of solute directly. However, we can find the molality of the new solution using the boiling point elevation formula: ∆Tb = Kb × m 1.78 °C = 2.34 °C·kg/mol × m m = 0.760 mol/kg Now we can calculate the moles of solute using the molality and the mass of CS₂: Moles of solute = molality × mass of solvent in kg Moles of solute = 0.760 mol/kg × 0.06305 kg = 0.04796 mol

Step 6: Calculate the molecular weight of the unknown solute

Now that we know the moles of the unknown solute, we can use its mass to find its molecular weight: Molecular weight = mass of solute / moles of solute Molecular weight = 5.39 g / 0.04796 mol = 112.4 g/mol The molecular weight of the unknown solute is 112.4 g/mol.

Key Concepts

Molality

Molality is an important concept in chemistry that describes the concentration of a solute in a solution. Unlike molarity, which is dependent on the volume of the solution, molality is based on the mass of the solvent.

Here's a simple way to think about it:

• Molality (\( m \)) is defined as the moles of solute divided by the mass of the solvent in kilograms.
• The formula for molality is: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

In the exercise example, for carbon disulfide (\( \mathrm{CS}_{2} \)), the molality is calculated by dividing the moles of the solute (given as 0.250 mol) by the kilograms of the solvent (calculated by taking the volume of \( \mathrm{CS}_{2} \) and multiplying by its density to find the mass, which is then converted to kilograms). The resulting calculation provides the molality, which is a crucial step in determining the boiling-point elevation of the solution.

Molal Boiling-Point-Elevation Constant

The molal boiling-point-elevation constant (\( K_b \)) is a specific constant that varies depending on the solvent used. This constant represents the boiling point elevation per molal concentration of the solution. Essentially, it tells us how much the boiling point of a solvent will increase when a certain amount of solute is added, per kilogram of solvent.

The change in boiling point (\( \Delta T_b \)) can be calculated using the formula:

• \[ \Delta T_b = K_b \times m \]
• Where \( \Delta T_b \) is the change in boiling point, \( K_b \) is the molal boiling-point-elevation constant, and \( m \) is the molality of the solution.

To find \( K_b \), in our example, we rearranged the formula as follows:

• \[ K_b = \frac{\Delta T_b}{m} \]

Using this formula, we plugged the values of the boiling-point elevation and the calculated molality to find \( K_b \) for carbon disulfide. This constant is critical for predicting how the boiling point changes with different solute concentrations.

Molecular Weight

Molecular weight is essential for identifying unknown compounds. It is calculated by dividing the mass of the substance by the amount of the substance in moles.

In our example, once we've determined the moles of an unknown solute by calculating its molality in the new solution, we can then find its molecular weight. The steps are as follows:

• Begin with the mass of the solute, given in the problem.
• Use the solved molality and the mass of the solvent to find the moles of solute.
• Finally, find the molecular weight with the formula:\[ \text{Molecular weight} = \frac{\text{mass of solute}}{\text{moles of solute}} \]

In the exercise, the unknown solute's mass was 5.39 grams and the determined moles was 0.04796 mol, leading to the calculation of the molecular weight as 112.4 g/mol. This value is crucial for characterizing and identifying chemical substances.