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Chapter 13: Problem 104

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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