Name Date Experiment 2 Advance Study Assignment: Resolution of Matter into Pure
ID: 873371 • Letter: N
Question
Name Date Experiment 2 Advance Study Assignment: Resolution of Matter into Pure Substances, I. Paper Chromatography 1. A student chromatographs observes the following: a mixture, and after developing the spots with a suitable reagent he tPoint of application Solvent front What are the R, values? 2. Explain, in your own words, why samples can often be separated into their components by The solvent moves 3 cm in about 10 minutes. Why shouldn't the experiment be stopped at that time instead of waiting 75 minutes for the solvent to move 10 cm? 3. In this experiment it takes the Cu(NO,, solution there in one spot? about 10 microliters of solution to produce a spot 1 em in diameter. If ), solution contains about 6 g Cu erliter, how many micrograms of Cu ion are 4. microgramsExplanation / Answer
1. Rf = distance of the spot from point of application/ total solvent eluted.
So,you have to measure the distance of the spot from the point of application using a scale and divide it with length L to get the corresponding Rf of a spot. For a particular spot, measure till the center of the spot.
2. In chromatography, a mixture can be separated to its corresponding componets because each has different Rf in a different solevnt. This is based on polarity of the components of the mixture. More polar, more high is the substance eluted or up in the chromatogram. So, it is very effective for substances with high difference in polarity.
3. generally, we wait for the solvent to rise till it raisen to 95% of the silica chromatogram that is L should be upto almost 95% of the silica paper. For your experiment, it take 75 mins. so u cannot stop at 3 minutes because the solvent will be raisen only 5-10% and you will not observe the separation of the components as the solvent has not eluted. So the compnents are not soaked to the solvent and risen up with it. If you stop at 3 minutes, the more polar components will not be shown on the plate and the separation will not be completed.
4.So the taken volume of solution is 10microliteres. Given that 1L has 6g Cu2+ , 1microliter has 6/(10^6) g Cu2+ = 6E-6 g Cu2+. So for 10microliters produce 6E-6 x 10 = 6E-5 g Cu2+
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