/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 What is the osmolarity of the fo... [FREE SOLUTION] | 91影视

91影视

What is the osmolarity of the following solutions? (a) \(0.35 M \mathrm{KBr}\) (b) \(0.15 M\) glucose \(+0.05 M \mathrm{~K}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(a) 0.70 osmoles/L; (b) 0.30 osmoles/L.

Step by step solution

01

Understanding Osmolarity

Osmolarity is the total concentration of all solute particles in a solution. It's calculated by adding the contribution of each solute's molarity multiplied by the number of particles it dissociates into.
02

Calculate for KBr

Potassium bromide (KBr) dissociates into two ions: K鈦 and Br鈦. Therefore, 1 mole of KBr will yield 2 moles of particles. The osmolarity for KBr is calculated as:\[0.35 \text{ M} \times 2 = 0.70 \text{ osmoles/L}\]
03

Calculate for Glucose

Glucose does not dissociate into ions in solution, so it remains as one particle. Therefore, the osmolarity contribution from glucose is the same as its molarity, which is 0.15 M.
04

Calculate for K2SO4

Potassium sulfate (K鈧係O鈧) dissociates into three ions: 2 K鈦 and 1 SO鈧劼测伝. Thus, 1 mole of K鈧係O鈧 yields 3 moles of particles. The osmolarity for K鈧係O鈧 is:\[0.05 \text{ M} \times 3 = 0.15 \text{ osmoles/L}\]
05

Calculate Total Osmolarity for Part (b)

Add the osmolarity contributions for glucose and K鈧係O鈧 together:\[0.15 \text{ osmoles/L (glucose)} + 0.15 \text{ osmoles/L (}\mathrm{K}_2\mathrm{SO}_4\text{)} = 0.30 \text{ osmoles/L}\]

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 91影视!

Key Concepts

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

Molarity
Molarity is a measure of the concentration of solute in a solution. It's defined as the number of moles of solute per liter of solution, abbreviated as "M". Understanding molarity helps in calculating how much of a substance is dissolved in a given volume of liquid.
For example, when we say a solution has a molarity of 0.35 M KBr, it means there are 0.35 moles of potassium bromide dissolved in every liter of solution. Molarity is a central concept in chemistry because it allows chemists to predict how a solution will behave in reactions, especially when it comes to mixing different solutions. Calculating molarity involves taking into account the total volume of the solution, which is vital for accurate measurements in laboratory settings.
To find the molarity, the formula is as follows:
  • Molarity (M) = Moles of solute / Liters of solution
This simple ratio helps us understand how concentrated a given solution is.
Dissociation
Dissociation refers to the process by which molecules split into smaller particles, such as ions, when dissolved in a solvent like water. It's a critical concept that explains how substances like salts and acids behave when they are mixed with water.
In the example exercise, KBr dissociates in water into K鈦 and Br鈦 ions. This means that when one mole of KBr dissolves, it produces 2 moles of ions in total. Understanding dissociation is essential for calculating osmolarity, as the number of particles each solute produces will influence the solution's properties.
Glucose, however, does not dissociate like ionic compounds. It remains as individual glucose molecules when dissolved, so its molarity and osmolarity are the same.
Ions
Ions are charged particles formed by the loss or gain of electrons. In solutions, ions interact with the solvent, which affects the solution's overall properties. They are the building blocks for understanding processes like conductivity and osmotic pressure.
When compounds like salts dissolve in water, they often dissociate into ions, which can carry an electric charge. For example, in our problem, KBr dissociates into K鈦 and Br鈦 ions. The presence of these ions in solution is what defines the ionic strength of the solution.
Ions are pivotal in calculating the osmolarity of a solution because each dissociated ion contributes individually to the total number of particles in the solution. As a rule of thumb, the more ions present, the higher the osmolarity of the solution.
Solution Concentration
Solution concentration quantifies the amount of solute present in a given quantity of solvent. It determines how strong or dilute a solution is, affecting how the solution will react chemically and physically.
In the context of our exercise, we considered how the concentration of each solute contributes to the total solution's osmolarity. By understanding the solutes' molarity and their dissociation behavior, we calculated the individual and total contributions to the solution concentration.
  • For KBr: Contribution is based on its molarity and dissociation into 2 ions.
  • For glucose: Contribution is based on its molarity only, as it does not dissociate.
  • For K鈧係O鈧: Contribution depends on it dissociating into 3 ions.
In summary, the concentration of solutions depends on both the amount of solute added and their subsequent behavior in the solution, which is why osmolarity can differ even for solutions with the same molarity.

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

Emergency treatment of cardiac arrest victims sometimes involves injection of a calcium chloride solution directly into the heart muscle. How many grams of \(\mathrm{CaCl}_{2}\) are administered in an injection of \(5.0 \mathrm{~mL}\) of a \(5.0 \%(\mathrm{~m} / \mathrm{v})\) solution? How many milliequivalents of \(\mathrm{Ca}^{2+}\) ?

Hard candy is prepared by dissolving pure sugar and flavoring in water and heating the solution to boiling. What is the boiling point of a solution produced by adding \(650 \mathrm{~g}\) of cane sugar (molar mass \(342.3 \mathrm{~g} / \mathrm{mol}\) ) to \(1.5 \mathrm{~kg}\) of water? (For each mole of nonvolatile solute, the boiling point of \(1 \mathrm{~kg}\) of water is raised \(0.51{ }^{\circ} \mathrm{C} / \mathrm{K} .\) )

The Environmental Protection Agency has set the limit for arsenic in drinking water at 0.010 ppm. To what volume would you need to dilute \(1.5 \mathrm{~L}\) of water containing 5.0 ppm arsenic to reach the acceptable limit?

Many compounds are only partially dissociated into ions in aqueous solution. Trichloroacetic acid \(\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right),\) for instance, is partially dissociated in water according to the equation $$ \mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}(a q) $$ For a solution prepared by dissolving \(1.00 \mathrm{~mol}\) of trichloroacetic acid in \(1.00 \mathrm{~kg}\) of water, \(36.0 \%\) of the trichloroacetic acid dissociates to form \(\mathrm{H}^{+}\) and \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\) ions. (a) What is the total concentration of dissolved ions and molecules in \(1 \mathrm{~kg}\) of water? (b) What is the freezing point of this solution? (The freezing point of \(1 \mathrm{~kg}\) of water is lowered \(1.86^{\circ} \mathrm{C} / \mathrm{K}\) for each mole of solute particles.)

The maximum amounts of lead and copper allowed in drinking water are \(0.015 \mathrm{mg} / \mathrm{kg}\) for lead and \(1.3 \mathrm{mg} / \mathrm{kg}\) for copper. Express these values in parts per million, and tell the maximum amount of each (in grams) allowed in \(100 \mathrm{~g}\) of water.

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.