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  1. Problems
  2. /A dye has an Rf of 0.42 in a solvent. The solvent front moves 11.5 cm…
MediumChemistryHigh School (9–10)GeneralMixtures and Solutions › Separation of MixturesNumerical

A dye has an Rf of 0.42 in a solvent. The solvent front moves 11.5 cm…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

  1. A dye has an RfR_fRf​ of 0.420.420.42 in a solvent. The solvent front moves 11.5 cm11.5\ \text{cm}11.5 cm. How far does the dye move?
  2. A second dye on the same paper has an RfR_fRf​ of 0.780.780.78. How far apart are the two spots?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

cm
cm
Need a reminder? The key ideaShowHide

In chromatography, each dye is carried up the paper at its own speed. Rf=distance moved by the spotdistance moved by the solventR_f = \dfrac{\text{distance moved by the spot}}{\text{distance moved by the solvent}}Rf​=distance moved by the solventdistance moved by the spot​, measured from the starting line. It has no unit, lies between 000 and 111, and identifies a substance in a given solvent.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. (a) d=Rf×D=0.42×11.5 cm=4.83 cmd = R_f \times D = 0.42 \times 11.5\ \text{cm} = 4.83\ \text{cm}d=Rf​×D=0.42×11.5 cm=4.83 cm.

Step 2. The second dye. 0.78×11.5 cm=8.97 cm0.78 \times 11.5\ \text{cm} = 8.97\ \text{cm}0.78×11.5 cm=8.97 cm.

Step 3. (b) The gap. 8.97 cm−4.83 cm=4.14 cm8.97\ \text{cm} - 4.83\ \text{cm} = 4.14\ \text{cm}8.97 cm−4.83 cm=4.14 cm.

Answer: (a) 4.83 cm4.83\ \text{cm}4.83 cm. (b) 4.14 cm4.14\ \text{cm}4.14 cm apart.

Check. Both spots are below the solvent front at 11.5 cm11.5\ \text{cm}11.5 cm, as they must be.

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