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Chapter 25: Q16P (page 708)

Microporous silica particles with a density of2.2g/mLand a diameter10μmof have a measured surface area of calculated the surface area of the300m2spherical silica as if it were simply solid particles what does this calculation tell you about the shape or porosity of the particles.

Short Answer

Expert verified

Microporous silica particles with a density calculation tell you about the shape or porosity of the particles.

First we calculate volume of sphere:

v=43.πr3v=43.π5.10-4cm3p=0.27m2

Step by step solution

01

definition of volume

The amount of three-dimensional space enclosed by a closed surface is expressed as a scalar quantity.

First we calculate volume of sphere:

v=43.πr3v=43.π5.10-4cm3p=5.24.10-4cm3

Then we calculate mass of one sphere:

m=ρ.vm=2.2g/mL.5.24.10-10mLm=1.15.10-9g

So, the number of particles in 1 g is:

role="math" localid="1654843608416" N=1g1.10.10-9gN=8.7.108

02

Step 2:surface one particle

The surface of one particle is:

p=4.πr2p=4.π5.10-6m2p=3.14.10-10m2

Now we calculate surface of particles:

p=3.10.10-10m2.8.7.102p=0.27m2

The measured surface area is 300m2, so the particles have irregular shapes and they are porous.

Hence,

p=0.27m2

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Most popular questions from this chapter

what are criteria for an adequate isocratic chromatographic separation?

A bonded stationary phase for the separation of optical isomers has the structure

To resolve the enantiomers of amines, alcohols, or thiols, the compounds are first derivatized with a nitroaromatic group that increases their interaction with the bonded phase and makes them observable with a spectrophotometric detector.

When the mixture is eluted with 20 vol% 2-propanol in hexane, the (R)-enantiomer is eluted before the (S)-enantiomer, with the following chromatographic parameters:

Resolution=Δtrwav=7.7

Relative retention(α)=4.53

k for (R)-isomer =1.35

tm=1.00min

wherewavis the average width of the two Gaussian peaks at their base.

(a) Findt1,t2andwawith units of minutes.

(b) The width of a peak at half-height isrole="math" localid="1663582749865" w1/2(Figure 23-9). If the plate number for cach peak is the same, findw1/2for each peak.

(c) The area of a Gaussian peak is1.064×peakheight×w1/2. Given that the areas under the two bands should be equal, find the relative peak heights(heightR//heightS).

If along 15cmHPCL column has a place height of 5.0 what will be the half-width (in seconds) of a peak eluted at 10.0min? if plate height5μm,what will bew1/2?

The figure shows the separation of two enantiomers on a chiral stationary phase.


Sepation of enantiomers of Ritalin by HPLC with a chiral stationary phase.[Data from R.Bakhitar,L.Ramos,and F.L.S.Tse, “Quantification of methlylphenidate in plasma using chiral Liquid-chromatography/Tandem mass spectrometry: Application to Toxicokinetric studies,”Anal Chim Acta 2002,469,261.]

(a)From trandw1/2 find N for each peak.

(b) Fromtrandw1/2find the resolution.

(c)Giventm=1.62min, use Equation23-23with the average N to predict the resolution.

(a) When you try separating an unknown mixture byreversed-phase chromatography with 48%acetonitrile 50%water,the peaks are too close together and are eluted in the range k = 2- 6Should you use a higher or lower concentration of acetonitrile in thenext run?

(b) When you try separating an unknown mixture by normal-phasechromatography with 50%hexane50% methyl t-butyl ether, thepeaks are too close together and are eluted in the range k = 2 - 6Should you use a higher or lower concentration of hexane in thenext run?
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