/*! 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} Q18P The partition coefficient for a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Q18P (page 629)

The partition coefficient for a solute in chromatography is K=cs/cm, where csis the concentration in the stationary phase and cmis the concentration in the mobile phase. The larger the partition coefficient, the longer it takes a solute to be eluted. Explain why.

Short Answer

Expert verified

concentration in stationary phase increases there is more concentration that needs to be extracted to the mobile phase and the elution would be longer.

Step by step solution

01

Given Information:

The Partition coefficient for a solute is K=cs/cm

02

Partition coefficient is larger :

The given equation is K=cs/cm

The concentration of the solute in the stationary phase is csproportional to the partition coefficient K

If a concentration in stationary phase increases there is more concentration that needs to be extracted to the mobile phase and the elution would be longer.

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!

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) A chromatography column with a length of 10.3 cmand inner diameter of 4.61mmis packed with a stationary phase that occupies 61.0% of the volume. If the volume flow rate is 1.13mL, find the linear velocity in cm/min.(b) How long does it take for solvent (which is the same as unretained solute) to pass through the column? (c) Find the retention time for a solute with a retention factor of 10.

23-47. A chromatographic peak has a width,W,of 4.0mLand a retention volume of role="math" localid="1654842133798" 49mL. What width is expected for a band with a retention volume of 127mL? Assume that the only band spreading occurs on the column itself.

Why is longitudinal diffusion a more serious problem in gas chromatography than in liquid chromatography?

The theoretical limit for extracting solute Sfrom Phase1(volumeV1)into phase2 (volumeV2)is attained by dividing V2into an infinite number of infinitesimally small portions and conducting an infinite number of extractions. With a partition coefficient, K=[S]2/[S]1the limiting fraction of solute remaining in phase 1is2qlimit=e-(V2/V1)K(V1=V2=50mLandK=2). Let volume V2be divided intoequal portions to conduct extractions. Find the fraction of S extracted into phase 2 for n = 1,2, and 10extractions. How many portions are required to attain 95%of the theoretical limit?

Match statements 1鈥�5 with the band broadening terms in the second list.

1. Depends on radius of open tubular column.

2. Not present in an open tubular column.

3. Depends on length and radius of connecting tubing.

4. Increases with diffusion coefficient of solute.

5. Increases with thickness of stationary phase film.

A Multiple paths

B Longitudinal diffusion

Cm Equilibration time in mobile phase

Cs Equilibration time in stationary phase

EC Extra column band broadening
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation
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.