How many colors would you see if you were analyzing a pure substance through chromatography?

For separation of components of mixture chromatography technique is used. Parts of mixture are separated by using the different solubilities of the parts for a stationary phase (that does not move) compared to the mobile phase (that carries the parts). Due to capillary action, the solvent will move up with paper or plate in apparatus.  The mobile phase will move with solvent and the stationary phase will remain adhere to the paper or plate. And thus, giving separate colors for the substances present in the mixture.

Hence, for pure substance, only one color would be observed through chromatography.

Answer :

The correct answer is  no color will observed if a pure substance is analyzed ,

Chromatography :

It is a technique commonly used to separate a mixture of chemical substances based on their affinity with stationary phase and mobile phase .

The  mobile phase travels through the stationary phase  in a definite direction . The component of mixture travels along with mobile according to their affinity , ionic interactions , diffusion , solubility  on stationary phase .

The component has more affinity with mobile phase goes higher with mobile phase .

While component with stationary phase remain at lower position  on stationary phase .

This is how separation of components takes place.

A pure substance has only one type of component in it . There is no mixture of components.

Hence when pure substance will be analyzed  by chromatography technique, there will be no separation of components .

So there will no color  be observed.  

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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