Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Raoult's Law
Raoult's Law is a principle used to determine the vapor pressure of solutions. It states that the vapor pressure of a solution is the product of the mole fraction of the solvent and the vapor pressure of the pure solvent. In a formula, this law is expressed as \( P = P^\circ \times X_1 \). Here, \( P \) represents the vapor pressure of the solution, \( P^\circ \) is the vapor pressure of the pure solvent, and \( X_1 \) is the mole fraction of the solvent. This concept helps in understanding how a solute affects the boiling point and vapor pressure when dissolved in a solvent.
To utilize Raoult's Law, it's important to accurately find the mole fraction of the solvent. This involves comparing the vapor pressures (both measured and of the pure state). Note however, this applies to ideal solutions where interactions between molecules are similar to the solvent alone.
Boiling Point
The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure. For liquids like ethanol, knowing the boiling point is crucial in calculating changes when solutes dissolve in it. At the boiling point, liquids turn to vapor. When a solute is added, it causes the boiling point to elevate, requiring a higher temperature for boiling.
In our example, ethanol's normal boiling point at one atmosphere pressure is 78.4°C, corresponding to a vapor pressure of 760 torr. Using Raoult's Law, the impact of adding non-volatile solute, such as the one in our problem, can be seen as it affects the solution's properties, increasing the boiling point.
Molar Mass Calculation
Calculating molar mass is an essential part of chemistry. It helps determine the amount of substance needed or produced in a reaction. Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol).
Following the steps, you use the molality derived from the mole fraction of the solute and the known mass of the solute. In the problem, the molar mass is found using the formula: Molar mass of solute = \( \frac{\text{mass of solute}}{\text{molality} \times \text{mass of solvent in kg}} \).
  • The molality is found through the mole fractions calculated using Raoult's Law.
  • Knowing the mass of the solute, you compute the molar mass.
  • In our context, the molar mass calculation confirms the solute's properties and identity as 348 g/mol.
This highlights the interplay between vapor pressure, boiling point, and molecular composition.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) 600 \(\mathrm{mL} .\) of 0.250 \(M \operatorname{SrBr}_{2},(\mathbf{b}) 86.4 \mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KCl},(\mathrm{c}) 124.0 \mathrm{g}\) of a solution that is 6.45\(\%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is 208 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\) . A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving 324 \(\mathrm{g}\) in 100 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

(a) Calculate the mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in a solution containing 10.6 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 483 \(\mathrm{g}\) of water. (b) An ore contains 2.86 \(\mathrm{g}\) of silver per ton of ore. What is the concentration of silver in ppm?

Calculate the molarity of the following aqueous solutions: (a) 0.540 \(\mathrm{g}\) of Mg \(\left(\mathrm{NO}_{3}\right)_{2}\) in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{b}) 22.4 \mathrm{gof}\) \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in 125 \(\mathrm{mL}\) of solution, \((\mathrm{c}) 25.0 \mathrm{mL}\) of 3.50 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to 0.250 \(\mathrm{L}\)

(a) A sample of hydrogen gas is generated in a closed container by reacting 2.050 g of zinc metal with 15.0 \(\mathrm{mL}\) . of 1.00 \(\mathrm{M}\) sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 122 mL. Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\) , ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.8 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) -atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free