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Chapter 13: Problem 102

The normal boiling point of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) is \(78.4^{\circ} \mathrm{C} .\) When 9.15 \(\mathrm{g}\) of a soluble nonelectrolyte is dissolved in 100.0 \(\mathrm{g}\) of ethanol at that temperature, the vapor pressure of the solution is \(7.40 \times 10^{2}\) torr. What is the molar mass of the solute?

Short Answer

Expert verified
The molar mass of the solute is 348 g/mol.

Step by step solution

01

Determine the molality of the solution

(using Raoult's Law) Raoult's Law states that the vapor pressure of a solution (P) is equal to the vapor pressure of the pure solvent (P°) times the mole fraction of the solvent (X₁): \[ P = P° * X₁ \]
02

Calculate the mole fraction of the solvent

Given: Vapor pressure of the solution (P) = \(7.40 \times 10^2\) torr Normal boiling point (Vapor pressure) of ethanol (P°) = 78.4°C (The normal boiling point is the temperature at which the vapor pressure of the substance in its liquid phase is equal to 1 atm or 760 torr) First, convert the normal boiling point to torr: P° = 78.4°C = (78.4°C + 273.15) K = 351.55 K Vapor pressure of pure ethanol = 760 torr (at its boiling point) So now we can find the mole fraction of the solvent (X₁ = P/P°): X₁ = P/P° = (\(7.40 \times 10^2\) torr) / (760 torr) = 0.9737
03

Calculate the mole fraction of the solute

Mole fraction of the solute (X₂) can be calculated as follows: X₂ = 1 - X₁ = 1 - 0.9737 = 0.0263
04

Calculate the molality of the solution

Molality (m) can be calculated as: m = X₂ * (1000 g/kg) / (mass of solvent in kg) Since we have 100 g of ethanol as solvent, we can convert it to kg: mass of solvent (ethanol) = 100.0 g = 0.1 kg Now calculate the molality: m = 0.0263 * (1000 / 0.1) = 0.0263 * 10000 = 263 mol/kg
05

Calculate the molar mass of the solute

Given that 9.15 g of solute is dissolved in 0.1 kg of ethanol, we can use the formula: Molar mass of solute = mass of solute / (molality * mass of solvent in kg) Molar mass of solute = 9.15 g / (263 mol/kg * 0.1 kg) = 9.15 g / 26.3 mol = 348 g/mol Hence, the molar mass of the solute is 348 g/mol.

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Most popular questions from this chapter

Oil and water are immiscible. Which is the most likely reason? (a) Oil molecules are denser than water. (b) Oil molecules are composed mostly of carbon and hydrogen. (c) Oil molecules have higher molar masses than water. (d) Oil molecules have higher vapor pressures than water. (e) Oil molecules have higher boiling points than water.

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 molar mass of the unknown?

The presence of the radioactive gas radon (Rn) in well water presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at \(30^{\circ} \mathrm{C}\) is \(7.27 \times 10^{-3} \mathrm{M},\) what is the Henry's law constant for radon in water at this temperature? (b) A sample consisting of various gases contains \(3.5 \times 10^{-6}\) mole fraction of radon. This gas at a total pressure of 32 atm is shaken with water at \(30^{\circ} \mathrm{C} .\) Calculate the molar concentration of radon in the water.

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C} ?\)

Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called "salting out" the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein-protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein's hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?
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