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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

Expert verified
The molal boiling-point-elevation constant for Carbon Disulfide (CS₂) is 2.34 °C/mol·kg. The molecular weight of the unknown solute is 257 g/mol.

Step by step solution

01

Calculate the boiling point elevation, ΔT

We have the boiling point of pure carbon disulfide and the boiling point of the solution. The boiling point elevation (ΔT) is the difference between these two values: ΔT = (47.46°C - 46.30°C) = 1.16°C
02

Calculate the molality of the solution, m

We have 0.250 mol solute in 400 mL carbon disulfide. First, we'll find the mass of carbon disulfide in grams: mass of CS₂ = volume × density mass of CS₂ = (400.0 mL) × (1.261 g/mL) = 504.4 g Now we can find the molality of the solution: molality, m = (moles of solute) / (kilograms of solvent) molality, m = (0.250 mol) / (504.4 g × 1 kg/1000 g) = 0.4957 mol/kg
03

Calculate K_b of CS_2

Now that we have the boiling point elevation (ΔT) and the molality of the solution (m), we can find K_b using the equation: ΔT = K_b × m Rearrange the equation to solve for K_b: K_b = ΔT / m Substitute the known values and solve: K_b = (1.16°C) / (0.4957 mol/kg) = 2.34 °C/mol·kg #b. Molecular weight of the unknown solute#
04

Calculate the boiling point elevation, ΔT, for the second solution

Again, we have the boiling point of the solution (47.08°C) and the boiling point of pure CS₂ (46.30°C). The boiling point elevation (ΔT) is the difference between these two values: ΔT = (47.08°C - 46.30°C) = 0.78°C
05

Calculate the molality of the second solution, m

Using the boiling point elevation equation that includes K_b, we can find the molality of the solution: ΔT = K_b × m Rearrange the equation to solve for molality, m: m = ΔT / K_b Substitute the known values and solve: m = (0.78°C) / (2.34 °C/mol·kg) = 0.333 mol/kg
06

Calculate the moles of the unknown solute

We are given the mass of the unknown solute (5.39 g) and the mass of the solvent (50.0 mL of CS₂). First, find the mass of the solvent in grams: mass of CS₂ = volume × density mass of CS₂ = (50.0 mL) × (1.261 g/mL) = 63.05 g molality, m = (moles of solute) / (kilograms of solvent) Rearrange the equation to solve for the moles of solute: moles of solute = molality × kilograms of solvent moles of solute = (0.333 mol/kg) × (63.05 g × 1 kg/1000 g) = 0.02098 mol
07

Calculate the molecular weight of the unknown solute

Now that we have the number of moles of solute and the mass of the solute, we can find the molecular weight of the unknown solute: molecular weight = (mass of solute) / (moles of solute) molecular weight = (5.39 g) / (0.02098 mol) = 257 g/mol The molecular weight of the unknown solute is 257 g/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molality
Molality is a measure of the concentration of a solute in a solvent. Unlike molarity, which depends on the volume of the solution, molality is based on the mass of the solvent. This makes it useful for calculations involving temperature changes, since mass doesn't change with temperature.

Here's how you determine molality:
  • Calculate the mass of the solvent in kilograms.
  • Use the formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)
For example, in the exercise, the mass of carbon disulfide is calculated by multiplying its volume by its density, giving 504.4 g or 0.5044 kg.
The molality is found by dividing the number of moles of solute by the mass of the solvent: \( m = \frac{0.250 \text{ mol}}{0.5044 \text{ kg}} \approx 0.4957 \text{ mol/kg} \).

This calculation is particularly handy when it comes to boiling point elevation.
Molecular Weight Calculation
Determining the molecular weight of an unknown substance can be exciting and is often done using the concept of boiling point elevation. It involves a few key steps:

Once the boiling point elevation (\( \Delta T \)) is known, and you have the molal boiling point elevation constant (\( K_b \)) for the solvent, you can find the molality (\( m \)) of the solution, as seen in the solution process. With the formula:
  • \( \Delta T = K_b \times m \)
You first solve for \( m \).

Using molality, calculate the moles of the unknown:
  • Moles of solute = molality \( \times \) kilograms of solvent.
The solution then reveals that the molality of the second solution is found by dividing the boiling point elevation by \( K_b \), yielding \( 0.333 \text{ mol/kg} \).

With these moles of solute, use the formula:
  • Molecular weight = \( \frac{\text{mass of solute}}{\text{moles of solute}} \)
This results in the specific molecular weight of the unknown solute, 257 g/mol.
Colligative Properties
Colligative properties are characteristics of solutions that depend on the number of solute particles, not their identity. This is an important concept as it explains why boiling point elevation happens.

There are several colligative properties, including:
  • Boiling Point Elevation: When a solute is dissolved in a solvent, the boiling point of the solvent increases. This happens because the solute particles interfere with the escape of solvent molecules into the vapor phase.
  • Freezing Point Depression: The freezing point of the solvent decreases when a solute is added because solute particles disrupt crystal formation.
  • Osmotic Pressure: Solvent flows through a semipermeable membrane from a dilute solution to a concentrated one.
For boiling point elevation, the equation used is:
\( \Delta T = K_b \times m \)
where \( \Delta T \) is the change in boiling point, \( K_b \) is the molal boiling point elevation constant, and \( m \) is the molality.

Using these colligative properties allows for the calculation of important solution behaviors and helps solve problems like finding molecular weights of unknown substances.

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Most popular questions from this chapter

Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) or glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) (Figure 13.12 ); (b) propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)\) or sodium propionate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\right) ;\) (c) HCl or ethyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\right) ?\) Explain in each case.

Two beakers are placed in a sealed box at \(25^{\circ} \mathrm{C}\). One beaker contains \(30.0 \mathrm{~mL}\) of a \(0.050 \mathrm{M}\) aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains \(30.0 \mathrm{~mL}\) of a \(0.035 \mathrm{M}\) aqueous solution of \(\mathrm{NaCl}\). The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?

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Indicate the type of solute-solvent interaction (Section 11.2\()\) that should be most important in each of the following solutions: (a) \(\mathrm{CCl}_{4}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),(\mathbf{b})\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water, (c) \(\mathrm{KBr}\) in water, (d) HCl in acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\).

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