Answer:
[tex] \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}[/tex]
Explanation:
In Analytical Chemistry chromatography is widely used for the separation of samples.
- In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
- The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
- Consider the mixture we have taken consist of two samples having large polar difference.
- Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
- The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.
In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.
Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.
Formula to calculate Rf value is
[tex] \sf R_f \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent} [/tex]
Now, solving for Rf value of sample 1
Given:
Distance moved by sample 1 = 11 mm
Distance movedby solvent = 35 mm
To find:
Rf value of sample 1 = ?
Solution:
Substituting the given data in above formula,
[tex] \small \sf R_f \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent} \\ \small \sf R_f \: value = \cancel\frac{11 \: mm}{35 \: mm} = 0.3142[/tex]
[tex] \small \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}[/tex]
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