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?

Short answer

(a) The molal boiling-point-elevation constant for Carbon disulfide (CS₂) is approximately \(2.34 ^\circ\text{C}·\text{kg/mol}\). (b) The molecular weight of the unknown compound is approximately \(257 \text{ g/mol}\).

Step-by-step solution

Calculate the change in boiling point

First, we need to find the change in boiling point (∆T) by subtracting the pure CS2 boiling point from the boiling point of the solution: ∆T = T_solution - T_pure ∆T = 47.46 °C - 46.30 °C ∆T = 1.16 °C

Calculate the molality of the solution

Next, we need to calculate the molality (m) of the solution. Molality is the number of moles of solute per kilogram of solvent. m = moles_solute / mass_solvent_in_kg We know that there are 0.250 mol of solute in the solution. We need to convert the volume of CS2 to mass, then to kilograms, using the given density. mass_CS2 = volume_CS2 × density_CS2 mass_CS2 = 400.0 mL × 1.261 g/mL mass_CS2 = 504.4 g Now, we convert the mass of CS2 into kilograms: mass_CS2_in_kg = 504.4 g × (1 kg / 1000 g) mass_CS2_in_kg = 0.5044 kg Now we can calculate the molality: m = 0.250 mol / 0.5044 kg m = 0.4957 mol/kg

Calculate the molal boiling-point-elevation constant

We can now calculate the molal boiling-point-elevation constant (K_b) using the formula: ∆T = K_b × m K_b = ∆T / m K_b = 1.16 °C / 0.4957 mol/kg K_b ≈ 2.34 °C·kg/mol (b) Finding the Molecular Weight of the Unknown Compound We will follow these steps: 1. Calculate the change in boiling point for the unknown compound. 2. Calculate the molality of the solution with the unknown compound. 3. Calculate the moles of the unknown compound. 4. Calculate the molecular weight of the unknown compound.

Calculate the change in boiling point for the unknown compound

First, we need to find the change in boiling point (∆T) as we did earlier. ∆T = T_solution - T_pure ∆T = 47.08 °C - 46.30 °C ∆T = 0.78 °C

Calculate the molality of the solution with the unknown compound

We know the K_b value from part (a), so we can use it to find the molality (m) of the solution: ∆T = K_b × m m = ∆T / K_b m = 0.78 °C / 2.34 °C·kg/mol m ≈ 0.333 mol/kg

Calculate the moles of the unknown compound

We can now calculate the moles of unknown compound using the molality and mass of the solvent: mol_unknown = m × mass_solvent_in_kg mass_solvent = 50.0 mL × 1.261 g/mL = 63.05 g mass_solvent_in_kg = 63.05 g × (1 kg / 1000 g) = 0.06305 kg mol_unknown = 0.333 mol/kg × 0.06305 kg mol_unknown ≈ 0.0210 mol

Calculate the molecular weight of the unknown compound

We can now calculate the molecular weight of the unknown compound using the moles and mass: Molecular_weight = mass_unknown / mol_unknown We are given that the mass of the unknown compound is 5.39 g. Molecular_weight = 5.39 g / 0.0210 mol Molecular_weight ≈ 257 g/mol The molecular weight of the unknown compound is approximately 257 g/mol.

Key concepts

Molality

Molality is a way to express the concentration of a solution. It is specifically useful in boiling-point elevation calculations because it does not change with temperature.
Molality (\( m \)) is the number of moles of solute per kilogram of solvent. Here’s how it works:

• First, determine the number of moles of solute, which is often given directly in a problem.
• Next, convert the solvent volume to mass using its density.
• Finally, convert the mass from grams to kilograms, then divide the moles of solute by the kilograms of solvent to find molality.

For example, in the original problem, we have 0.250 moles of a solute in 0.5044 kg of CS₂, which gives a molality of 0.4957 mol/kg.
This information allows us to determine other thermodynamic properties, especially related to temperature changes in solutions.

Molecular Weight

The molecular weight of a substance is a fundamental property that tells us how much one mole of the compound weighs. It is the sum of the atomic masses of all atoms in a molecule, expressed in grams per mole. Molecular weight is crucial in various scientific calculations, including determining the moles of a substance when given its mass.
When tasked to find the molecular weight of an unknown compound, here is a straightforward approach:

• Calculate the change in boiling point when the solute is added.
• Use the known molal boiling-point-elevation constant (\( K_b \)) and the measured change in boiling point (\( \Delta T \)) to find the molality of the solution.
• Determine the number of moles using the molality and mass of the solvent.
• Finally, calculate the molecular weight by dividing the mass of the unknown compound by the number of moles.

In our example, an unknown compound's mass was 5.39 g, with 0.0210 moles calculated, resulting in a molecular weight of approximately 257 g/mol.

Solution Density

Solution density is the ratio of the mass of a solution to its volume, typically expressed in grams per milliliter (g/mL). It is an essential parameter when converting between volume and mass, which is particularly helpful in molality calculations where mass is required.To calculate the mass of a given volume of a solution:

• Multiply the volume (in mL) by the density of the solution. This gives the mass in grams.
• Convert that mass to kilograms by dividing by 1000.

Using the density of carbon disulfide (\( 1.261 \, \text{g/mL} \)) from the problem, you can find the mass of the solvent like this:For the first part, 400.0 mL of CS₂ yields 504.4 g. Converting to kilograms results in 0.5044 kg.
Understanding solution density helps clarify the relationship between different concentration measures and is vital in performing accurate calculations in chemistry.