/*! 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} Q36P Consider the chromatography of n... [FREE SOLUTION] | 91影视

91影视

Consider the chromatography of n-c12H26on a 25 m 脳 0.53 mm open tubular column of 5% phenyl鈥95% methyl polysiloxane with a stationary phase thickness of 3.0 渭尘and He carrier gas at 125潞C. The observed retention factor for n-c12H26is 8.0. Measurements were made of plate height, H, at various values of linear velocity,xm/s. A least-squares curve through the data points is given by

localid="1654864230381" H(m)=(6.010-5m2/s)/x+(2.0910-3s)xthecoefficientsofthevanDeemterequation,findthediffusioncoefficientofn-c12H26inthemobileandstationaryphases.Whyisoneofthesediffusioncoefficientssomuchgreaterthantheother?

Short Answer

Expert verified

Fromsubstitutingthevalueswegetthediffusioncoefficientofn-C12H26inmobilephaseis3.010-5m2sandthediffusioncoefficientofn-C12H16instationaryphaseis5.010-10m2/s.ThediffusioncoefficientofmobileisgreaterthanthestationaryphasebecauseitiseasierforthediffusionofsolutethroughHeliumgasthanthroughaviscousliquidphase.

Step by step solution

01

The deemter equation is taken in the form

H=Bx+Cx=(Cs+Cm)xCs=2k3k+12d2dsCm=1+6k11k224k+12r2DmB=2Dm

WhereDm=diffusioncoefficientofsoluteinmobilephasek=retentionfactord=thicknessofstationaryphaseDs=diffusioncoefficientofsoluteinstationaryphase

02

Diffusion coefficient of solute in mobile phase

H=6.010-5m2/sx+(2.0910-3s)xB=2DM=6.010-5m2/sDm=3.0105m2/sForthesecondterm,Cs+Cm=2.0910-3s=2k3(k+1)2d2Ds+1+6k+11k224(k+1)2r2dmconsumptionofcitrateandactivationofglycolysis.

03

Diffusion coefficient of solute in stationary phase

Substituting the values of all parameters to solve Ds

2.0910-3s=28.038.0+12(3.010-6m)2Ds+1+6(8.0)+11(80)224((8.0)+1)2(2.6510-4m)2(3.010-5m2/s)Ds=5.010-10m2/s

04

Diffusion coefficient in mobile phase

Diffusion coefficient in mobile phase=3.010-5m2/s5.010-10m2/s=6.0104

05

Reason why one of these diffusion coefficient is greater than the other

Diffusion coefficient in mobile phase is 6.010-4times higher than the diffusion coefficient in stationary phase. The diffusion coefficient of mobile is greater than the diffusion coefficient in the stationary phase because it is easier for the diffusion of solute through Helium gas than through liquid phase.

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

Heptane, decane, and an unknown had adjusted retention times of 12.6min(heptane),22.9min(decane), and20.0min(unknown). The retention indexes for heptane and decane are 700 and 1000 , respectively. Find the retention index for the unknown.

This problem reviews concepts from Chapter 23 using

Figure 24-7.

(a) Calculate the number of theoretical plates (N in Equation 23-30)

and the plate height (H) for CO.

(b) Find the resolution (Equation 23-23) between argon and oxygen.

(a) Why is it illogical to use a thin stationary phase (0.2 碌m) in a wide-bore (0.53-mm) open tubular column?

(b) Consider a narrow-bore (0.25 mm diameter), thin-film (0.10 碌m) column with 5 000 plates per meter. Consider also a wide-bore (0.53 mm diameter), thick-film (5.0 碌m) column with 1500 plates per meter. The density of stationary phase is approximately 1.0 g/mL. What mass of stationary phase is in each column in a length equivalent to one theoretical plate? How many nanograms of analyte can be injected into each column if the mass of analyte is not to exceed 1.0% of the mass of stationary phase in one theoretical plate?

The antitumor drug gimatecan is available as nearly pure (S)-enantiomer. Neither pure (R)-enantiomer nor a racemic (equal) mixture of the two enantiomers is available. To measure small quantities of (R)-enantiomer in nearly pure (S)-gimatecan, a preparation was subjected to normal-phase chromatography on each of the enantiomers of a commercial, chiral stationary phase designated (S,S)- and (R,R)-DACH-DNB. Chromatography on the (R,R)-stationary phase gave a slightly asymmetric peak at tr 5 6.10 min with retention factor k 5 1.22. Chromatography on the (S,S)- stationary phase gave a slightly asymmetric peak at tr 5 6.96 min with k 5 1.50. With the (S,S) stationary phase, a small peak with 0.03% of the area of the main peak was observed at 6.10 min.

Chromatography of gimatecan on each enantiomer of a chiral stationary phase. Lower traces have enlarged vertical scale. [Data from E. Badaloni, W. Cabri, A. Ciogli, R. Deias, F. Gasparrini, F. Giorgi, A. Vigevani, and C. Villani, 鈥淐ombination of HPLC 鈥業nverted Chirality Columns Approach鈥 and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.鈥 Anal. Chem. 2007, 79, 6013.]

(a) Explain the appearance of the upper chromatograms. Dashed lines are position markers, not part of the chromatogram. What Problems 709 would the chromatogram of pure (R)-gimatecan look like on the same two stationary phases?

(b) Explain the appearance of the two lower chromatograms and why it can be concluded that the gimatecan contained 0.03% of the (R)-enantiomer. Why is the (R)-enantiomer not observed with the (R,R)-stationary phase?

(c) Find the relative retention (a) for the two enantiomers on the (S,S)-stationary phase.

(d) The column provides N 5 6 800 plates. What would be the resolution between the two equal peaks in a racemic (equal) mixture of (R)- and (S)-gimatecan? If the peaks were symmetric, does this resolution provide baseline separation in which signal returns to baseline before the next peak begins?

This problem reviews concepts from Chapter 23 using

Figure 24-15.

(a) Calculate the retention factor for peak 11 given tm 5 6.7 min.

(b) Calculate the number of theoretical plates (N in Equation 23-31)

and the plate height (H) for peak 11.

(c) Find the resolution (Equation 23-23) between peaks 16 and 17.

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.