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

An aqueous solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) having a concentration of \(34.2\) gram/litre has an osmotic pressure of \(2.38\) atmospheres at \(17^{\circ} \mathrm{C}\). For an aqueous solution of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to be isotonic with this solution, its concentration should be : (a) \(34.2\) gram per litre (b) \(17.1\) gram per litre (c) \(18.0\) gram per litre (d) \(36.0\) gram per litre

Short Answer

Expert verified
The concentration for the glucose solution should be 18.0 grams per litre (Option c).

Step by step solution

01

Understand the Concept of Isotonic Solutions

Isotonic solutions have the same osmotic pressure. This means that the concentration of solute particles must be the same. Since both sucrose and glucose dissociate into different numbers of particles, their respective molar concentrations will determine if the solutions are isotonic.
02

Calculate the Molar Concentration of the Sucrose Solution

To find the molar concentration (M) of sucrose, use the formula M = mass / (molar mass * volume). Given a mass of 34.2 g/L and the molar mass of sucrose as 342 g/mol, the molar concentration is M_sucrose = 34.2 / 342 mol/L.
03

Calculate the Required Concentration for the Glucose Solution

Since the solutions are isotonic, the molar concentrations must be equal, M_sucrose = M_glucose. With the molar mass of glucose as 180 g/mol, the concentration in grams per liter for the glucose solution is found by multiplying the molar concentration from Step 2 by the molar mass of glucose, concentration_glucose = M_glucose * 180 g/mol.
04

Solve for the Concentration of Glucose

Using the previous steps, the glucose concentration is concentration_glucose = (34.2 / 342) * 180 g/L.

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.

Osmotic Pressure
Osmotic pressure is a fundamental concept in chemistry, particularly when discussing solutions and their properties. It is defined as the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane, which separates the solution from the pure solvent. The flow of solvent occurs due to osmosis, the movement of solvent molecules from an area of lower solute concentration to an area of higher solute concentration to achieve equilibrium.

The osmotic pressure of a solution is directly related to its molar concentration and the temperature of the solution. This relationship is described by the formula: \( \Pi = iMRT \), where \( \Pi \) represents the osmotic pressure, \( i \) is the van 't Hoff factor (which accounts for the number of particles the solute dissociates into), \( M \) is the molar concentration, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. In the context of isotonic solutions, they must have the same osmotic pressure—thus if two solutions are isotonic, their osmotic pressures are equal, ensuring no net movement of solvent across the semipermeable membrane separating them.
Molar Concentration
Molar concentration, commonly referred to as molarity, is a measure of the concentration of a solute in a solution. It is expressed in moles of solute per liter of solution, and its unit is moles per liter (mol/L). To calculate the molar concentration, one can use the formula: \( M = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{volume of solution}} \).

For instance, in the provided exercise, the molar concentration of a sucrose solution was calculated using the mass of sucrose and its molar mass. Understanding the concept of molarity is critical when trying to make two solutions isotonic, as we need to match their molar concentrations in order to ensure that their osmotic pressures are equivalent. Once the molar concentration of one solution is known, the same concentration can be used to find out the mass of the second solute (such as glucose) needed to achieve an isotonic state.
Colligative Properties
Colligative properties are characteristics of solutions that depend on the ratio of solute particles to solvent particles, irrespective of the identities of the solute particles. These properties are influenced by the number of particles dissolved in a given amount of solvent and include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

Knowing about colligative properties is vital because they imply that any non-volatile solute that dissociates in the solvent can affect the solution's overall behavior. When two solutions are isotonic, one of the colligative properties—osmotic pressure—is equal between them, which means they have the same number of solute particles per unit of solvent. An important aspect of colligative properties is that they allow us to predict and control the conditions under which certain reactions and processes will occur, which is crucial in fields like chemistry, medicine, and engineering.

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

A complex is represented as \(\mathrm{CoCl}_{3} x \mathrm{NH}_{3}\). Its \(0.1\) molal solution in water shows \(\Delta T_{f}=0.558 \mathrm{~K}\). \(K_{f}\) for \(\mathrm{H}_{2} \mathrm{O}\) is \(1.86 \mathrm{~K}\) molality \(^{-1}\). Assuming \(100 \%\) ionisation of complex and co-ordination number of Co is six, calculate formula of complex: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (d) none of these

If \(M_{\text {normal }}\) is the normal molecular mass and \(\alpha\) is the degree of ionization of \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), then the abnormal molecular mass of the complex in the solution will be: (a) \(M_{\text {normal }}(1+2 \alpha)^{-1}\) (b) \(M_{\text {normal }}(1+3 \alpha)^{-1}\) (c) \(M_{\text {normal }}(1+\alpha)^{-1}\) (d) equal to \(M_{\text {normal }}\)

The van't Hoff factor \(i\) for an electrolyte which undergoes dissociation and association in solvent are respectively: (a) greater than one and less than one (b) less than one and greater than one (c) less than one and less than one (d) greater than one and greater than one

Formation of a solution from two components can be considered as : (i) Pure solvent \(\rightarrow\) separated solvent molecules, \(\Delta H_{1}\) (ii) Pure solute \(\rightarrow\) separated solute molecules, \(\Delta H_{2}\) (iii) separated solvent and solute molecules \(\rightarrow\) solution, \(\Delta H_{3}\) Solution so formed will be ideal if : (a) \(\Delta H_{\text {Soln }}=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}\) (b) \(\Delta H_{\text {Soln }}=\Delta H_{1}+\Delta H_{2}-\Delta H_{3}\) (c) \(\Delta H_{\text {Soln }}=\Delta H_{1}-\Delta H_{2}-\Delta H_{3}\) (d) \(\Delta H_{\text {Soln }}=\Delta H_{3}-\Delta H_{1}-\Delta H_{2}\)

At 760 torr pressure and \(20^{\circ} \mathrm{C}\) temperature, \(1 \mathrm{~L}\) of water dissolves \(0.04 \mathrm{gm}\) of pure oxygen or \(0.02 \mathrm{gm}\) of pure nitrogen. Assuming that dry air is composed of \(20 \%\) oxygen and \(80 \%\) nitrogen (by volume). The masses (in \(\mathrm{g} / \mathrm{L}\) ) of oxygen and nitrogen dissolved by \(1 \mathrm{~L}\) of water at \(20^{\circ} \mathrm{C}\) exposed to air at a total pressure of 760 torr are respectively : (a) \(0.008,0.016\) (b) \(0.016,0.008\) (c) \(0.16,0.08\) (d) \(0.04,0.02\)

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