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

\(6.0 \mathrm{~g}\) of urea (molecular weight \(=60\) ) was dissolved in \(9.9\) moles of water. If the vapour pressure of pure water is \(P^{\circ}\), the vapour pressure of solution is: (a) \(0.10 P^{\circ}\) (b) \(1.10 P^{\circ}\) (c) \(0.90 P^{\circ}\) (d) \(0.99 P^{\circ}\)

Short Answer

Expert verified
The vapour pressure of the solution is (d) \(0.99 P^{\circ}\).

Step by step solution

01

Calculate the number of moles of urea

To find the number of moles of urea, use the formula: \(\text{Number of moles} = \frac{\text{Mass of substance}}{\text{Molecular weight}}\). So, for urea: \( \text{Number of moles} = \frac{6.0 \text{ g}}{60 \frac{\text{g}}{\text{mol}}} = 0.1 \text{ mol} \).
02

Determine the total number of moles in solution

Combine the moles of urea and water to get the total number of moles in the solution: \(0.1 \text{ mol urea} + 9.9 \text{ mol water} = 10.0 \text{ mol total}\).
03

Calculate the mole fraction of water

The mole fraction of water is the ratio of the number of moles of water to the total number of moles in the solution: \( \text{Mole fraction of water} = \frac{9.9}{10.0} = 0.99 \).
04

Calculate the vapour pressure of the solution

Apply Raoult's Law, which states that the vapour pressure of a solution is proportionate to the mole fraction of the solvent: \( P_{\text{solution}} = \text{Mole fraction of water} \times P^{\circ} = 0.99 \times P^{\circ} \).

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
When diving into the understanding of solution chemistry, especially when it comes to vapor pressure, Raoult's Law serves as a foundational principle. It elegantly connects the macroscopic property of vapor pressure to the composition of a mixture on the molecular level. Simply put, Raoult's Law states that the partial vapor pressure of a solvent in a solution is directly proportional to the solvent's mole fraction in the solution.

In the exercise provided, we're applying Raoult's Law to determine the vapor pressure of an aqueous urea solution. The law formula can be expressed as: \[ P_{\text{solution}} = (\text{Mole fraction of the solvent}) \times P^{\circ} \], where \(P^{\circ}\) is the vapor pressure of the pure solvent (water, in this case). For the steps mentioned, after determining the mole fraction of water, we multiply it by the pure water's vapor pressure to arrive at the solution's vapor pressure.
Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of one component to the total number of moles of all components in the mixture. The mole fraction is unitless and provides a measure of the proportion of a substance within a mixture.

Calculating the mole fraction is an essential step in the exercise. For example, the mole fraction of water in our urea solution is calculated by dividing the number of moles of water by the total number of moles in the solution (the sum of moles of water and urea). Understanding mole fraction is crucial not only for applications of Raoult's Law but also for many other calculations involving solutions, such as determining concentrations and colligative properties.
Colligative Properties
Colligative properties are properties of solutions that depend solely on the number of dissolved particles in the solution, not on the nature of the chemical species present. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

Understanding colligative properties helps us predict how a solution will behave compared to its pure solvent. In the context of vapor pressure, Raoult's Law describes how the presence of a solute (like urea) will lower the vapor pressure of the solvent (water). This lowering is a colligative property because it is directly related to the number of urea particles in the solution, regardless of their identity. The more solute particles present, the more the vapor pressure is lowered.
Solution Chemistry
Solution chemistry is an area of study that explores the interactions between solutes and solvents to form solutions. It encompasses understanding solubility, concentration measures (such as mole fraction), and the various colligative properties. In solution chemistry, the role of intermolecular forces is discussed to explain why substances dissolve and how the properties of the solution differ from those of the pure substances.

In our specific exercise, solution chemistry is at play when dissolving urea in water. The solute (urea) interacts with the solvent (water), affecting physical properties such as the vapor pressure. By studying solution chemistry, we can predict these changes and understand the underlying molecular interactions that drive them.

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

Which one of the following aqueous solutions will exhibit highest boiling point: (a) \(0.015 \mathrm{M}\) urea (b) \(0.01 \mathrm{M} \mathrm{KNO}_{3}\) (c) \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.015 \mathrm{M}\) glucose

Consider \(0.1 \mathrm{M}\) solutions of two solutes \(X\) and \(Y\). The solute \(X\) behaves as a univalent electrolyte while the solute \(Y\) dimerises in solution. Which of the following statements are correct regarding these solutions? (1) The boiling point of the solution of \(X\) will be higher than that of \(Y\) (2) The osmotic pressure of the solution of \(Y\) will be lower than that of \(X\) (3) The freezing point of the solution of \(X\) will be lower than that of \(Y\) (4) The relative lowering of vapour pressure of both the solutions will be the same Select the correct answer from the option given below : (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,2 and 4 (d) 1,3 and 4

Which of the following solutions will have maximum osmotic pressure? Assume \(90 \%\) dissociation of each salt: (a) decinormal aluminium sulphate (b) decinormal barium chloride solution (c) decinormal sodium sulphate solution (d) solution of equal volumes of decinormal barium chloride and decinormal sodium sulphate solutions

Four solutions of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) with the concentrations \(0.1 \mathrm{~m}, 0.01 \mathrm{~m}, 0.001 \mathrm{~m}\) and \(0.0001 \mathrm{~m}\) are available. The maximum value of colligative property corresponds to : (a) \(0.0001 \mathrm{~m}\) solution (b) \(0.001 \mathrm{~m}\) solution (c) \(0.01 \mathrm{~m}\) solution (d) \(0.1 \mathrm{~m}\) solution

\(0.1 \mathrm{M} \mathrm{NaCl}\) and \(0.05 \mathrm{M} \mathrm{BaCl}_{2}\) solutions are separated by a semi-permeable membrane in a container. For this system, choose the correct answer: (a) There is no movement of any solution across the membrane (b) Water flows from \(\mathrm{BaCl}_{2}\) solution towards \(\mathrm{NaCl}\) solution (c) Water flows from \(\mathrm{NaCl}\) solution towards \(\mathrm{BaCl}_{2}\) solution (d) Osmotic pressure of \(0.1 \mathrm{M} \mathrm{NaCl}\) is lower than the osmotic pressure of \(\mathrm{BaCl}_{2}\) (assume complete dissociation)

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