Why is it advantageous to use large particles \(\left( {{\bf{50}}{\rm{ }}\mu {\bf{m}}} \right)\) for solid phase extraction, but small particles \(\left( {{\bf{5}}{\rm{ }}\mu {\bf{m}}} \right)\) for chromatography?

Short Answer

Expert verified
  • If you're working with an aqueous base sample and want to inject it into a GC instrument to convert it to a volatile organic, solid phase extraction could help you achieve your aim.
  • Chromatography helps in purification,precise separation and analysis. It necessitates very small sample volumes.

Step by step solution

01

Definition ofChromatography.

  • Chromatography is a fashion for separating the factors, or solutes, of a mixture grounded on the relative quantities of each solute distributed between a flowing fluid sluice, known as the mobile phase, and a stationary phase that's conterminous.
02

Determine the advantages for solid particles and small particles for chromatography.

Solid phase extraction:

  • As per the physical and chemical properties the solid phase extraction helps in separating compounds dissolved or suspended in a liquid solution from other compounds in the mixture
  • Without application of high-pressure large particles can be allowed for the sample to drain through the column.

Chromatography

  • The separation of mixtures whilst using a mobile phase (moving solvent) on stationary phase\( \to \)mostly due to different polarity of each compound in the mixture.
  • Small particles - cause an increase in the efficiency of the separation. Using of these molecules requirehigh-pressure application in order to "push" the solvent through the column.
  • It can analyse pharmaceuticals, food particles, plastics, pesticides, air and water samples, and tissue extracts, among other things.

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Most popular questions from this chapter

To pre-concentrate cocaine and benzoylecgonine from river water described at the opening of thischapter, solid-phase extraction was carried out at2mL pH2using the mixed-mode cationexchange resin in Figure 28-19. After passing 500mLof river water through 60mgof resin, the retained analytes were eluted first with2mLof localid="1663594337127" CH3OHand then with localid="1663594104084" 2mLof2% ammonia solution inCH3OH. Explain the purpose of using pH2for retention and dilute ammonia for elution.

An example of a mixture of 1-mm-diameter particles of \({\rm{KCl}}\)and \({\rm{KN}}{{\rm{O}}_3}\)in a number ratio \(1:99\)follows Equation 28-4. A sample containing \({10^4}\)particles weighs\(11.0\;{\rm{g}}\). What is the expected number and relative standard deviation of \({\rm{KCl}}\)particles in a sample weighing\(11.0 \times {10^2}\;{\rm{g}}\)?

Explain how to prepare a powder with an average particle diameter near \(100\mu m\) by using sieves from Table 28-2. How would such a particle mesh size be designated?

Question: Consider a random mixture containing \(4.00\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) (density\(2.532g/mL\)) and \(96.00\;{\rm{g}}\)of \({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\) (density\(2.428\;{\rm{g}}/{\rm{mL}}\)) with a uniform spherical particle radius of\(0.075\;{\rm{mm}}\).

(a) Calculate the mass of a single particle of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and the number of particles of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) in the mixture. Do the same for\({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\).

(b) What is the expected number of particles in \(0.100\;{\rm{g}}\)of the mixture?

(c) Calculate the relative sampling standard deviation in the number of particles of each type in a \(0.100\;{\rm{g}}\)sample of the mixture.

When you flip a coin, the probability of its landing on each side is \(p = q = \frac{1}{2}\)in Equations 28-2 and 28-3. If you flip it \(n\)times, the expected number of heads equals the expected number of tails \( = np = nq = \frac{1}{2}n.\)The expected standard deviation for \(n\)flips is\({\sigma _n} = \sqrt {npq} \). From Table 4-1, we expect that \(68.3\% \)of the results will lie within \( \pm 1{\sigma _n}\) and \(95.5\% \)of the results will lie within\( \pm 2{\sigma _n}\).

(a) Find the expected standard deviation for the number of heads in \({\bf{1000}}\) coin flips.

(b) By interpolation in Table 4-1, find the value of \(z\)that includes \(90\% \)of the area of the Gaussian curve. We expect that \(90\% \)of the results will lie within this number of standard deviations from the mean.

(c) If you repeat the\({\bf{1000}}\)coin flips many times, what is the expected range for the number of heads that includes\(90\% \) of the results? (For example, your answer might be, "The range \({\bf{490}}\) to \({\bf{510}}\) will be observed \(90\% \)of the time.")

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